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Discrete random variables

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52380512
A spinner has three equal sections labeled \(1\), \(2\), and \(3\). The spinner is spun twice. The random variable \(S\) is the sum of the two results. a) Create a probability distribution table for \(S\). b) Find the probability that the sum is at least \(4\).

Hints

- List the ordered pairs that can result from two spins. - Determine how many equally likely outcomes there are. - Organize the outcomes in a list or grid to find each possible sum. - Translate “at least” into an inequality before selecting outcomes.

Solution

1. Each result has probability \(\frac{1}{3}\), so the two spins have \(3\cdot3=9\) equally likely ordered outcomes: \((1, 1)\), \((1, 2)\), \((1, 3)\), \((2, 1)\), \((2, 2)\), \((2, 3)\), \((3, 1)\), \((3, 2)\), and \((3, 3)\). 2. Count the outcomes for each sum: \(S=2\) occurs once, \(S=3\) twice, \(S=4\) three times, \(S=5\) twice, and \(S=6\) once. Therefore, the probabilities are \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{2}{9}\), and \(\frac{1}{9}\), respectively. 3. For a sum of at least \(4\), add the probabilities for \(S=4\), \(S=5\), and \(S=6\): \(P(S\ge4)=\frac{3}{9}+\frac{2}{9}+\frac{1}{9}=\frac{6}{9}=\frac{2}{3}\).

Answer

a) <table border="1"> <tr> <td>\(s\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> <td>\(6\)</td> </tr> <tr> <td>\(P(S=s)\)</td> <td>\(\frac{1}{9}\)</td> <td>\(\frac{2}{9}\)</td> <td>\(\frac{3}{9}\)</td> <td>\(\frac{2}{9}\)</td> <td>\(\frac{1}{9}\)</td> </tr> </table> b) \(P(S\ge4)=\frac{2}{3}\approx66.7\%\)
52511312
Determine whether each random variable is discrete or continuous. Briefly justify each answer by describing its possible values. a) The time for a \(100\,\text{m}\) sprint, assuming unlimited measurement precision. b) The number of defective circuit boards in a batch of \(1000\). c) The exact tire pressure of a car, measured in \(\text{bar}\). d) The sum obtained when three standard number cubes are rolled at the same time.

Hints

- Decide whether the variable can take any value in an interval or only separated values. - Can the possible outcomes be counted? - Does measurement precision change the theoretical type of the variable? - Distinguish a count from a physical measurement.

Solution

1. a) Sprint time is continuous because it can take any real value within a reasonable interval. 2. b) The number of defective boards is discrete because its possible values are the integers in \(\{0,1,\ldots,1000\}\). 3. c) Tire pressure is continuous because, in theory, it can take any real value within an interval. 4. d) The sum is discrete because it can take only the integer values from \(3\) through \(18\).

Answer

a) continuous b) discrete c) continuous d) discrete
52511412
A city planning office collects data about local development. Decide whether a discrete or continuous model is appropriate for each random variable. a) The number of residents registered in each apartment building. b) The exact area of a parcel of land in square meters. c) A household's monthly electric bill, recorded to the nearest cent. d) A visitor's waiting time at a city service office when the time is recorded by rounding down to a whole number of minutes.

Hints

- Imagine marking the possible values on a number line. Are they separated points or a continuous interval? - Pay attention to rounding and minimum units such as one cent. - Is the set of possible recorded values countable, or does it contain every real value in an interval?

Solution

1. a) The number of residents is discrete because people are counted in whole numbers. 2. b) Land area is continuous because an area measurement can theoretically take any real value within an interval. 3. c) The electric bill is discrete as recorded because the possible values differ by \(\$0.01\). 4. d) The recorded waiting time is discrete because rounding down produces only whole-number values \(\{0,1,2,\ldots\}\), even though the underlying waiting time is continuous.

Answer

a) discrete b) continuous c) discrete d) discrete
52514112
One probability model describes the number of children per household in a city. A second model describes residents' commuting times to work, measured theoretically with exact precision in minutes. Explain why a bar graph is appropriate for the first probability distribution, while a probability density curve is appropriate for the second. In your explanation, discuss the sets of possible values.

Hints

- Can all possible numbers of children be listed individually? - What happens to the possible commuting times as measurement precision increases? - How are probabilities represented for discrete and continuous random variables?

Solution

1. The number of children is a discrete random variable. Its possible values are separated, countable numbers such as \(0,1,2,\ldots\). Each value can be assigned a probability \(P(X=k)\), which can be represented by the height of a bar. 2. Exact commuting time is a continuous random variable. It can take any real value in an appropriate interval, so its set of possible values is uncountable. 3. For a continuous random variable, \(P(X=x)=0\) at every single value. Interval probabilities are represented by areas under a density curve, making a probability density function appropriate.

Answer

The number of children is discrete and has separated, countable values, so individual probabilities can be shown as bar heights. Exact commuting time is continuous. Since individual values have probability \(0\), interval probabilities are represented by areas under a density curve.
53085312
An irregular four-sided die has faces labeled \(1\), \(2\), \(3\), and \(4\). The table gives part of the probability distribution of the random variable \(X\), the number rolled. <table> <tr> <td>\(k\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.18\)</td> <td>\(0.32\)</td> <td>\(p\)</td> <td>\(0.25\)</td> </tr> </table> 1. Determine the probability \(p\). 2. Find the probability that the number rolled is odd. 3. Find the probability that the number rolled is at least \(2\).

Hints

- Recall the required sum of all probabilities in a distribution. - Identify the odd face values. - Translate “at least” into an inequality. - Consider using a complement for the last part.

Solution

1. The probabilities must add to \(1\): \(0.18+0.32+p+0.25=1\). Therefore, \(0.75+p=1\), so \(p=0.25\). 2. The odd outcomes are \(1\) and \(3\). Thus, \(P(X\in\{1,3\})=0.18+0.25=0.43\). 3. The outcomes at least \(2\) are \(2\), \(3\), and \(4\). Thus, \(P(X\ge2)=0.32+0.25+0.25=0.82\). Equivalently, \(P(X\ge2)=1-P(X=1)=1-0.18=0.82\).

Answer

1. \(p=0.25\) 2. \(P(\text{odd})=0.43\) 3. \(P(X\ge2)=0.82\)
53096112
An urn contains five balls labeled \(1\) through \(5\). Two balls are drawn in order without replacement. The random variable \(X\) is the absolute difference between the two numbers drawn. Determine the set of all possible values of \(X\).

Hints

- List pairs of different numbers that could be drawn. - Find the absolute difference for each pair. - Identify the smallest and largest possible differences, then check whether every integer between them can occur.

Solution

1. Because the balls are drawn without replacement, the two numbers must be different. 2. The smallest possible absolute difference is \(|1-2|=1\), and the largest is \(|1-5|=4\). 3. Each integer difference between them is possible: \(|1-2|=1\), \(|1-3|=2\), \(|1-4|=3\), and \(|1-5|=4\). 4. Therefore, the set of possible values is \(\{1, 2, 3, 4\}\).

Answer

\(\{1, 2, 3, 4\}\)
53097712
A customer service center recorded visitor wait times. The random variable \(X\) is the wait time, in minutes, for a randomly selected visitor. The table gives cumulative probabilities. <table> <thead> <tr> <th>Wait time \(x\), in minutes</th> <th>\(P(X\le x)\)</th> </tr> </thead> <tbody> <tr><td>\(2\)</td><td>\(0.18\)</td></tr> <tr><td>\(5\)</td><td>\(0.45\)</td></tr> <tr><td>\(10\)</td><td>\(0.72\)</td></tr> <tr><td>\(15\)</td><td>\(0.88\)</td></tr> <tr><td>\(20\)</td><td>\(0.96\)</td></tr> <tr><td>\(30\)</td><td>\(1.00\)</td></tr> </tbody> </table> Use the table to determine each probability. a) A visitor waits more than \(10\) minutes. b) \(P(5<X\le20)\) c) A visitor waits more than \(2\) minutes but no more than \(10\) minutes.

Hints

- Interpret each table entry as a cumulative probability. - Relate “more than \(x\)” to the complement of “at most \(x\).” - To find a probability between two cutoffs, subtract the cumulative value at the lower cutoff from the value at the upper cutoff. - Identify which visitors are included up to the upper cutoff but not up to the lower cutoff.

Solution

1. Use the complement: \(P(X>10)=1-P(X\le10)=1-0.72=0.28\). 2. Subtract cumulative probabilities: \(P(5<X\le20)=P(X\le20)-P(X\le5)=0.96-0.45=0.51\). 3. Similarly, \(P(2<X\le10)=P(X\le10)-P(X\le2)=0.72-0.18=0.54\).

Answer

a) \(P(X>10)=0.28\) b) \(P(5<X\le20)=0.51\) c) \(P(2<X\le10)=0.54\)
53115512
A spinner has five equal sections labeled \(1\), \(1\), \(2\), \(5\), and \(10\). The spinner is spun once. The random variable \(X\) assigns each section its displayed number. 1. Give the sample space \(S\) and the set of possible values of \(X\). 2. Create a probability distribution table for \(X\). 3. Briefly explain how to construct a probability histogram for this distribution and what each bar height represents when every bar has width \(1\).

Hints

- Distinguish the physical spinner sections from the numerical values assigned to them. - Count how many sections display each value. - Check that all probabilities add to \(1\). - Recall which quantities appear on the axes of a probability histogram.

Solution

1. Treat the five physical sections as distinguishable outcomes, so \(S=\{s_1,s_2,s_3,s_4,s_5\}\). The set of possible values is \(\{1,2,5,10\}\). 2. Two of the five sections show \(1\), so \(P(X=1)=\frac{2}{5}=0.4\). Each other value appears once, so \(P(X=2)=P(X=5)=P(X=10)=\frac{1}{5}=0.2\). 3. Place the possible values of \(X\) on the x-axis and probabilities on the y-axis. Draw one width-\(1\) bar at each possible value. The height, and therefore the area, of each bar equals \(P(X=k)\).

Answer

1. \(S=\{s_1,s_2,s_3,s_4,s_5\}\); the possible values are \(\{1,2,5,10\}\). 2. <table> <tr><td>\(k\)</td><td>\(1\)</td><td>\(2\)</td><td>\(5\)</td><td>\(10\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.4\)</td><td>\(0.2\)</td><td>\(0.2\)</td><td>\(0.2\)</td></tr> </table> 3. Draw a width-\(1\) bar at each possible value. Its height and area equal the corresponding probability.
54724912
A discrete random variable \(X\) takes values \(1,2,3,4\). Its cumulative distribution function satisfies \(F(1)=0.10\), \(F(2)=0.40\), \(F(3)=0.85\), and \(F(4)=1\). Recover the probability mass function and verify that it is valid.

Hints

- A discrete CDF accumulates all mass at or below each value. - Successive differences isolate individual point probabilities. - Check monotonicity and the final total.

Solution

1. \(P(X=1)=F(1)=0.10\). 2. \(P(X=2)=F(2)-F(1)=0.30\). 3. \(P(X=3)=F(3)-F(2)=0.45\). 4. \(P(X=4)=F(4)-F(3)=0.15\). 5. The probabilities are nonnegative and sum to \(1\).

Answer

\(P(X=1)=0.10\), \(P(X=2)=0.30\), \(P(X=3)=0.45\), and \(P(X=4)=0.15\).
54727212
Let \(I_A\) and \(I_B\) be the indicator variables of events \(A\) and \(B\). Explain why \(A\) and \(B\) are mutually exclusive exactly when \(I_AI_B=0\) for every outcome. What value would \(I_AI_B\) take on an outcome in the overlap?

Hints

- Recall when an event indicator equals \(1\). - Consider all four possible pairs of indicator values. - Connect the product being \(1\) with simultaneous occurrence.

Solution

1. An indicator equals \(1\) precisely when its event occurs. 2. The product \(I_AI_B\) equals \(1\) exactly when both indicators equal \(1\), which is exactly on \(A\cap B\). 3. Therefore the product is identically \(0\) exactly when the intersection is empty. 4. On an overlapping outcome, \(I_AI_B=1\).

Answer

Mutual exclusivity is equivalent to \(I_AI_B=0\) for every outcome. On an outcome in \(A\cap B\), the product equals \(1\).
54729912
A random variable \(X\) has cumulative distribution function \(F\), with \(F(1)=0.25\) and \(F(4)=0.70\). The real line is partitioned into the disjoint events \(E_1=\{X\le1\}\), \(E_2=\{1<X\le4\}\), and \(E_3=\{X>4\}\). Find the probability of each event and verify that the three probabilities sum to \(1\).

Hints

- A CDF gives the probability to the left of a cutoff. - The middle interval is obtained by subtracting two cumulative probabilities. - The final interval is the complement of the cumulative event at \(4\).

Solution

1. \(P(E_1)=F(1)=0.25\). 2. \(P(E_2)=F(4)-F(1)=0.70-0.25=0.45\). 3. \(P(E_3)=1-F(4)=0.30\). 4. The events are disjoint and exhaustive, and \(0.25+0.45+0.30=1\).

Answer

\(P(E_1)=0.25\), \(P(E_2)=0.45\), and \(P(E_3)=0.30\). Their sum is \(0.25+0.45+0.30=1\).
54739812
A discrete random variable \(X\) can take the values \(0,1,2,3\). Its probability function is \(P(X=x)=c(2x+1)\) for \(x\in\{0,1,2,3\}\). a) Find \(c\). b) Determine whether an odd value of \(X\) is more likely than an even value of \(X\). Support your answer with probabilities.

Hints

- A valid probability distribution must assign a total probability of one. - First compare the unscaled weights attached to the possible values. - For the comparison, group the outcomes according to whether they are odd or even.

Solution

1. The probability weights are \(1,3,5,7\), and their sum is \(16\). 2. Since the probabilities must total \(1\), \(16c=1\), so \(c=\frac{1}{16}\). 3. The odd-value probability is \(P(X=1)+P(X=3)=\frac{3}{16}+\frac{7}{16}=\frac{5}{8}\). 4. The even-value probability is \(P(X=0)+P(X=2)=\frac{1}{16}+\frac{5}{16}=\frac{3}{8}\), so an odd value is more likely.

Answer

a) \(c=\frac{1}{16}\) b) An odd value is more likely because \(P(X\text{ is odd})=\frac{5}{8}\) and \(P(X\text{ is even})=\frac{3}{8}\).
54740112
A random integer from \(00\) through \(99\) is used to simulate a discrete random variable \(X\) with the following distribution. <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.18\)</td><td>\(0.27\)</td><td>\(0.41\)</td><td>\(0.14\)</td></tr> </table> A student proposes this assignment: - \(00\)–\(17\) represents \(X=0\). - \(18\)–\(44\) represents \(X=1\). - \(45\)–\(86\) represents \(X=2\). - \(87\)–\(99\) represents \(X=3\). a) Explain why the assignment is incorrect and correct it by moving just one random integer to a different outcome. b) Using the target distribution, find \(P(X\ge2)\).

Hints

- Count the integers in each inclusive range carefully. - Each two-digit random integer represents the same fraction of the simulation outcomes. - Compare the counts assigned to the last two values with their target probabilities.

Solution

1. The proposed ranges contain \(18\), \(27\), \(42\), and \(13\) integers, respectively, so the last two probabilities would be \(0.42\) and \(0.13\), not \(0.41\) and \(0.14\). 2. Move \(86\) from the range for \(X=2\) to the range for \(X=3\). The corrected ranges are \(45\)–\(85\) and \(86\)–\(99\). 3. The required probability is \(P(X\ge2)=0.41+0.14=0.55\).

Answer

a) The proposed ranges give \(42\) integers to \(X=2\) and only \(13\) integers to \(X=3\). Assign \(45\)–\(85\) to \(X=2\) and \(86\)–\(99\) to \(X=3\); the first two ranges stay unchanged. b) \(P(X\ge2)=0.55\)
54740212
The letters in \(AABC\) are arranged in a uniformly random order. Let \(X\) be the number of letters between the two \(A\)'s. a) Construct the probability distribution of \(X\). b) Find the probability that the two \(A\)'s are not adjacent.

Hints

- Focus on the pair of positions occupied by the repeated letter. - Count how many position pairs produce each possible separation. - Translate “not adjacent” into a condition on the random variable.

Solution

1. The two positions occupied by the \(A\)'s are equally likely to be any of the \(\binom{4}{2}=6\) pairs of positions. 2. Three position pairs are adjacent, two have one position between them, and one has two positions between them. 3. Therefore, \(P(X=0)=\frac{3}{6}=\frac{1}{2}\), \(P(X=1)=\frac{2}{6}=\frac{1}{3}\), and \(P(X=2)=\frac{1}{6}\). 4. The two \(A\)'s are not adjacent when \(X>0\), so the probability is \(\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{6}\)</td></tr> </table> b) \(P(X>0)=\frac{1}{2}\)
54740612
The random variable \(X\) is the number of revisions needed for a randomly selected technical diagram. Its probability distribution is shown. <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(4\)</th><th>\(7\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.08\)</td><td>\(0.24\)</td><td>\(0.31\)</td><td>\(0.25\)</td><td>\(0.12\)</td></tr> </table> Define the \(75\)th percentile as the smallest value \(q\) for which \(P(X\le q)\ge0.75\). a) Find the \(75\)th percentile of \(X\). b) Find \(P(X<q)\) for that value of \(q\).

Hints

- Build cumulative totals from the smallest possible value upward. - Look for the first value where the accumulated probability reaches the stated threshold. - Pay attention to the difference between a strict inequality and an inclusive one.

Solution

1. The cumulative probabilities at \(x=0,1,2,4,7\) are \(0.08,0.32,0.63,0.88,1.00\), respectively. 2. The first cumulative probability that reaches or exceeds \(0.75\) occurs at \(x=4\), so \(q=4\). 3. The probability strictly below \(4\) is \(0.08+0.24+0.31=0.63\).

Answer

a) The \(75\)th percentile is \(q=4\). b) \(P(X<4)=0.63\)
54740812
Two distinct cards are selected at random from cards numbered \(1\) through \(5\). Let \(X\) be the smaller of the two numbers. a) Construct the probability distribution of \(X\). b) Find \(P(X\ge3)\).

Hints

- Treat the unordered pairs as equally likely outcomes. - For a proposed minimum, count which larger card can accompany it. - Check that every pair is counted exactly once.

Solution

1. There are \(\binom{5}{2}=10\) equally likely pairs. 2. A minimum of \(1\) occurs with any of the other four cards; a minimum of \(2\) occurs with one of \(3,4,5\); a minimum of \(3\) occurs with \(4\) or \(5\); and a minimum of \(4\) occurs only with \(5\). 3. Therefore, the probabilities for \(X=1,2,3,4\) are \(\frac{4}{10},\frac{3}{10},\frac{2}{10},\frac{1}{10}\). 4. Thus, \(P(X\ge3)=\frac{2}{10}+\frac{1}{10}=\frac{3}{10}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{2}{5}\)</td><td>\(\frac{3}{10}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{1}{10}\)</td></tr> </table> b) \(P(X\ge3)=\frac{3}{10}\)
54741012
An integer \(N\) is selected uniformly at random from \(1\) through \(12\). Let \(X=|N-6|\), the distance between the selected integer and \(6\). a) Construct the probability distribution of \(X\). b) Find \(P(X\ge4)\).

Hints

- Evaluate the distance from \(6\) for each possible selected integer. - Group integers that produce the same value of the random variable. - Convert each group count into a probability out of \(12\).

Solution

1. For \(N=1,2,\ldots,12\), the corresponding values of \(X\) are \(5,4,3,2,1,0,1,2,3,4,5,6\). 2. The value \(0\) occurs once, each of \(1,2,3,4,5\) occurs twice, and \(6\) occurs once. 3. Therefore, \(P(X=0)=P(X=6)=\frac{1}{12}\), and \(P(X=x)=\frac{1}{6}\) for \(x=1,2,3,4,5\). 4. Thus, \(P(X\ge4)=\frac{1}{6}+\frac{1}{6}+\frac{1}{12}=\frac{5}{12}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{12}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{12}\)</td></tr> </table> b) \(P(X\ge4)=\frac{5}{12}\)
54741112
The numbers \(1,2,3\) are arranged in a uniformly random order. Let \(X\) be the number of adjacent pairs that increase from left to right. a) Construct the probability distribution of \(X\). b) Find the probability that exactly one adjacent pair increases.

Hints

- List the possible orders of the three numbers. - Examine both neighboring comparisons in each order. - Group arrangements by the number of increasing comparisons.

Solution

1. There are \(3!=6\) equally likely arrangements. 2. The arrangement \(321\) has \(0\) increases, \(123\) has \(2\) increases, and each of the other four arrangements has exactly \(1\) increase. 3. Therefore, \(P(X=0)=\frac{1}{6}\), \(P(X=1)=\frac{4}{6}=\frac{2}{3}\), and \(P(X=2)=\frac{1}{6}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{6}\)</td><td>\(\frac{2}{3}\)</td><td>\(\frac{1}{6}\)</td></tr> </table> b) \(P(X=1)=\frac{2}{3}\)
54741212
Two distinct numbers are selected uniformly at random from \(\{2,3,4,6\}\). Let \(X\) be the positive difference between the larger and smaller selected numbers. a) Construct the probability distribution of \(X\). b) Find the probability that \(X\) is even.

Hints

- List the unordered pairs of selected numbers. - Find the positive difference for each pair. - Group pairs that produce the same difference before identifying the even values.

Solution

1. The \(\binom{4}{2}=6\) equally likely pairs produce differences \(1,2,4,1,3,2\). 2. Thus, \(X=1\) occurs for \(2\) pairs, \(X=2\) for \(2\) pairs, and \(X=3\) and \(X=4\) each occur for \(1\) pair. 3. Therefore, \(P(X=1)=\frac{1}{3}\), \(P(X=2)=\frac{1}{3}\), and \(P(X=3)=P(X=4)=\frac{1}{6}\). 4. The even values are \(2\) and \(4\), so \(P(X\text{ is even})=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{6}\)</td></tr> </table> b) \(P(X\text{ is even})=\frac{1}{2}\)
54741312
A simulation will draw one chip uniformly at random from a bag of \(24\) chips. Each chip must be labeled \(0\), \(2\), or \(5\). The simulated random variable \(X\) must satisfy \(P(X=0)=\frac{1}{8}\), \(P(X=2)=\frac{5}{12}\), and \(P(X=5)=\frac{11}{24}\). a) Determine how many chips should receive each label. b) Find \(P(X\ge2)\).

Hints

- Convert each probability into a count out of the total number of equally likely chips. - Verify that the resulting counts use every chip exactly once. - Identify which labels satisfy the inequality in part b.

Solution

1. Multiply each target probability by \(24\), the number of equally likely chips. 2. The required counts are \(24\cdot\frac{1}{8}=3\), \(24\cdot\frac{5}{12}=10\), and \(24\cdot\frac{11}{24}=11\). 3. These counts total \(24\), so they define the required simulation. 4. The event \(X\ge2\) uses labels \(2\) and \(5\), giving \(\frac{10+11}{24}=\frac{7}{8}\).

Answer

a) Label \(3\) chips with \(0\), \(10\) chips with \(2\), and \(11\) chips with \(5\). b) \(P(X\ge2)=\frac{7}{8}\)
54741712
Three distinct integers are selected uniformly at random from \(\{1,2,3,4,5\}\). Let \(X\) be the number of odd integers selected. a) Construct the probability distribution of \(X\). b) Find the probability that odd integers are a majority of the selection.

Hints

- Separate the available integers into odd and even groups. - Count selections with each possible number of odd integers. - Translate “majority” into a condition on the count.

Solution

1. There are \(\binom{5}{3}=10\) equally likely selections, with \(3\) odd and \(2\) even integers available. 2. Exactly one odd integer occurs in \(\binom{3}{1}\binom{2}{2}=3\) selections. 3. Exactly two odd integers occur in \(\binom{3}{2}\binom{2}{1}=6\) selections, and exactly three occur in \(\binom{3}{3}\binom{2}{0}=1\) selection. 4. Therefore, the probabilities for \(X=1,2,3\) are \(\frac{3}{10},\frac{3}{5},\frac{1}{10}\). 5. Odd integers are a majority when \(X\ge2\), giving \(\frac{3}{5}+\frac{1}{10}=\frac{7}{10}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{3}{10}\)</td><td>\(\frac{3}{5}\)</td><td>\(\frac{1}{10}\)</td></tr> </table> b) \(P(X\ge2)=\frac{7}{10}\)
54741812
A digit \(D\) is selected uniformly at random from \(0\) through \(9\). Let \(X\) be the last digit of \(D^2\). a) Construct the probability distribution of \(X\). b) Find \(P(X>5)\).

Hints

- Evaluate the rule for each equally likely input digit. - Record only the final digit of each square. - Group input digits that produce the same output value.

Solution

1. The last digits of \(D^2\) for \(D=0,1,\ldots,9\) are \(0,1,4,9,6,5,6,9,4,1\). 2. The values \(0\) and \(5\) each occur once; \(1,4,6,9\) each occur twice. 3. Therefore, \(P(X=0)=P(X=5)=\frac{1}{10}\), and \(P(X=1)=P(X=4)=P(X=6)=P(X=9)=\frac{1}{5}\). 4. The values greater than \(5\) are \(6\) and \(9\), so \(P(X>5)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(9\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{10}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{1}{10}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{1}{5}\)</td></tr> </table> b) \(P(X>5)=\frac{2}{5}\)
54742012
Two red cards and two blue cards are arranged in a uniformly random order. Let \(X\) be the number of times the color changes between neighboring cards. a) Construct the probability distribution of \(X\). b) Find the probability that \(X\) is even.

Hints

- List distinct color arrangements rather than treating same-colored cards as distinguishable. - Check each neighboring pair in an arrangement. - Group arrangements by the resulting number of changes.

Solution

1. The six equally likely color arrangements are \(RRBB,RBBR,RBRB,BBRR,BRRB,BRBR\). 2. Two arrangements have \(1\) color change, two have \(2\) color changes, and two have \(3\) color changes. 3. Therefore, \(P(X=1)=P(X=2)=P(X=3)=\frac{1}{3}\). 4. The only even possible value is \(2\), so \(P(X\text{ is even})=\frac{1}{3}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{3}\)</td></tr> </table> b) \(P(X\text{ is even})=\frac{1}{3}\)
54742312
Determine which of the following rules define valid discrete probability distributions. Justify each decision. a) For \(x\in\{-1,0,1\}\), the probabilities are \(0.20,0.50,0.30\), respectively. b) For \(x\in\{1,2,3\}\), \(P(X=x)=\frac{x}{5}\). c) For \(x\in\{0,1,2\}\), \(P(X=x)=\frac{(x-1)^2}{2}\). If rule c) is valid, state the values that have positive probability.

Hints

- Check nonnegativity and the total probability for each rule separately. - Evaluate formula-defined probabilities at every listed value. - A listed value may receive probability zero without making the rule invalid.

Solution

1. Rule a) has nonnegative probabilities and \(0.20+0.50+0.30=1\), so it is valid. 2. Rule b) has total probability \(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}=\frac{6}{5}\), so it is invalid. 3. Rule c) gives probabilities \(\frac{1}{2},0,\frac{1}{2}\), which are nonnegative and total \(1\), so it is valid. 4. Under rule c), only \(0\) and \(2\) have positive probability.

Answer

a) Valid. b) Invalid because the probabilities total \(\frac{6}{5}\). c) Valid; the values with positive probability are \(0\) and \(2\).
54742412
A four-character binary string is selected uniformly at random from all strings containing exactly two \(1\)'s and two \(0\)'s. Let \(X\) be the length of the longest consecutive block of \(0\)'s. Construct the probability distribution of \(X\), and find \(P(X=1)\).

Hints

- List only strings with the required number of each symbol. - For each string, inspect every block of consecutive zeros. - Group the strings according to the longest block found.

Solution

1. The six equally likely strings are \(0011,0101,0110,1001,1010,1100\). 2. The strings \(0101,0110,1010\) have longest zero block \(1\). 3. The strings \(0011,1001,1100\) have longest zero block \(2\). 4. Therefore, \(P(X=1)=\frac{3}{6}=\frac{1}{2}\) and \(P(X=2)=\frac{1}{2}\).

Answer

<table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{2}\)</td></tr> </table> \(P(X=1)=\frac{1}{2}\)
54742512
Two distinct vertices of the regular polygon shown are selected uniformly at random. Let \(X\) be the smaller number of sides along the perimeter needed to travel from one selected vertex to the other. a) Construct the probability distribution of \(X\). b) Find \(P(X\ge2)\).
Figure for problem 547425

Hints

- Count unordered pairs of vertices. - Classify each pair by its shortest separation around the perimeter. - Use the symmetry of the polygon to count pairs in each class.

Solution

1. The polygon has \(6\) vertices, so there are \(\binom{6}{2}=15\) equally likely vertex pairs. 2. Six pairs are adjacent and give \(X=1\); six pairs have one vertex between them and give \(X=2\); three pairs are opposite and give \(X=3\). 3. Therefore, \(P(X=1)=\frac{2}{5}\), \(P(X=2)=\frac{2}{5}\), and \(P(X=3)=\frac{1}{5}\). 4. Thus, \(P(X\ge2)=\frac{2}{5}+\frac{1}{5}=\frac{3}{5}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{2}{5}\)</td><td>\(\frac{2}{5}\)</td><td>\(\frac{1}{5}\)</td></tr> </table> b) \(P(X\ge2)=\frac{3}{5}\)
54742612
An integer \(B\) is selected uniformly at random from \(\{-3,-2,-1,0,1,2,3\}\). Let \(X\) be the number of positive values among \(B-1\) and \(B+1\). a) Construct the probability distribution of \(X\). b) Find the probability that at least one of the two values is positive.

Hints

- Evaluate \(B-1\) and \(B+1\) for each possible value of \(B\). - Count how many of the two results are positive in each case. - Group the seven equally likely values of \(B\) by the resulting count.

Solution

1. For \(B=-3,-2,-1\), neither \(B-1\) nor \(B+1\) is positive, so \(X=0\) in \(3\) cases. 2. For \(B=0,1\), exactly one of the two values is positive, so \(X=1\) in \(2\) cases. 3. For \(B=2,3\), both values are positive, so \(X=2\) in \(2\) cases. 4. Therefore, \(P(X=0)=\frac{3}{7}\), \(P(X=1)=\frac{2}{7}\), and \(P(X=2)=\frac{2}{7}\). 5. At least one value is positive when \(X\ge1\), so \(P(X\ge1)=\frac{4}{7}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{3}{7}\)</td><td>\(\frac{2}{7}\)</td><td>\(\frac{2}{7}\)</td></tr> </table> b) \(P(X\ge1)=\frac{4}{7}\)
54742712
One cell is selected uniformly at random from the grid shown. Let \(X\) be the number of other cells that share a full side with the selected cell. a) Construct the probability distribution of \(X\). b) Find \(P(X\ge3)\).
Figure for problem 547427

Hints

- Separate the cells into corner, edge, and center types. - Determine the number of side-sharing neighbors for each type. - Convert the number of cells of each type into probabilities.

Solution

1. The grid has \(9\) cells. Each of the \(4\) corner cells shares a side with \(2\) cells. 2. Each of the \(4\) noncorner edge cells shares a side with \(3\) cells. 3. The center cell shares a side with \(4\) cells. 4. Therefore, \(P(X=2)=\frac{4}{9}\), \(P(X=3)=\frac{4}{9}\), and \(P(X=4)=\frac{1}{9}\). 5. Thus, \(P(X\ge3)=\frac{4}{9}+\frac{1}{9}=\frac{5}{9}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{4}{9}\)</td><td>\(\frac{4}{9}\)</td><td>\(\frac{1}{9}\)</td></tr> </table> b) \(P(X\ge3)=\frac{5}{9}\)
54742912
One nonempty subset of \(\{A,B,C,D\}\) is selected uniformly at random from all nonempty subsets. Let \(X\) be the number of elements in the selected subset. a) Construct the probability distribution of \(X\). b) Find the probability that \(X\) is even.

Hints

- Count all subsets first, remembering that the empty subset is excluded. - Group subsets by their size. - Use the subset counts for the even sizes in part b.

Solution

1. There are \(2^4-1=15\) nonempty subsets. 2. The numbers of subsets of sizes \(1,2,3,4\) are \(\binom{4}{1}=4\), \(\binom{4}{2}=6\), \(\binom{4}{3}=4\), and \(\binom{4}{4}=1\). 3. Therefore, the probabilities are \(\frac{4}{15},\frac{2}{5},\frac{4}{15},\frac{1}{15}\). 4. An even size is \(2\) or \(4\), so \(P(X\text{ is even})=\frac{6+1}{15}=\frac{7}{15}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{4}{15}\)</td><td>\(\frac{2}{5}\)</td><td>\(\frac{4}{15}\)</td><td>\(\frac{1}{15}\)</td></tr> </table> b) \(P(X\text{ is even})=\frac{7}{15}\)
52512312
A manufacturer produces metal rods. The length \(L\), in centimeters, of a rod is modeled as a continuous normal random variable with \(L\sim N(100,0.1^2)\). The number \(X\) of defective rods in a daily production run of \(500\) is modeled as a discrete binomial random variable with \(p=0.02\). 1. Explain the main difference between the sets of possible values for \(L\) and \(X\). 2. Find \(P(L=100)\). 3. Find \(P(X=10)\). 4. Explain mathematically why a continuous random variable such as \(L\) always has probability \(0\) at an exact value, while a discrete random variable such as \(X\) may have positive probability at an exact value.

Hints

- Compare a finite list of integers with an unbroken interval of real numbers. - What is the area under a curve over an interval of width \(0\)? - Recall the binomial probability formula. - Compare probability represented by area with probability assigned to individual outcomes.

Solution

1. The possible values of \(X\) are the finite set \(\{0,1,2,\ldots,500\}\). The normal random variable \(L\) can take any real value in its model. 2. For any continuous random variable, \(P(L=x)=0\) at a single value, so \(P(L=100)=0\). 3. Using the binomial probability formula, \(P(X=10)=\binom{500}{10}(0.02)^{10}(0.98)^{490}\approx 0.1264\). 4. For a continuous random variable, probability is area under a density curve: \(P(a\le L\le b)=\int_a^b f(x)\,dx\). When \(a=b\), the interval has width \(0\), so the integral is \(0\). A discrete probability distribution can assign positive probability directly to individual values.

Answer

1. \(X\) has finitely many separated values, while \(L\) can take any real value in its model. 2. \(P(L=100)=0\) 3. \(P(X=10)\approx 0.1264\) 4. A continuous probability is an area over an interval, and a single point has zero width; a discrete distribution can assign positive probability to an individual value.
52512412
Consider two random experiments. Experiment A: \(Y\) is the sum when two fair six-sided number cubes are rolled. Experiment B: \(Z\) is normally distributed with \(\mu=7\) and \(\sigma=2.4\). 1. Classify \(Y\) and \(Z\) as discrete or continuous, and justify each classification. 2. Find \(P(Y\le 3)\). 3. Find \(P(6.5\le Z\le 7.5)\). 4. Explain how including endpoints, such as comparing \(P(X\le k)\) with \(P(X<k)\), affects probabilities for each type of random variable.

Hints

- List the possible sums from rolling two number cubes. - Use standardization or a normal cumulative distribution function for \(Z\). - What area does one isolated point contribute under a density curve? - How many ordered rolls produce a sum of at most \(3\)?

Solution

1. \(Y\) is discrete because its possible values are the integers \(\{2,3,\ldots,12\}\). \(Z\) is continuous because a normal random variable can take any real value. 2. The outcomes with \(Y\le 3\) are \((1, 1)\), \((1, 2)\), and \((2, 1)\). There are \(6\cdot 6=36\) equally likely ordered outcomes, so \(P(Y\le 3)=\frac{3}{36}=\frac{1}{12}\approx 0.0833\). 3. Standardize the endpoints: \(z_1=\frac{6.5-7}{2.4}\approx -0.2083\) and \(z_2=\frac{7.5-7}{2.4}\approx 0.2083\). Therefore, \(P(6.5\le Z\le 7.5)=\Phi(0.2083)-\Phi(-0.2083)\approx 0.1650\). 4. For discrete \(Y\), an endpoint may have positive probability, so \(P(Y\le 3)=P(Y<3)+P(Y=3)\). For continuous \(Z\), \(P(Z=k)=0\), so including or excluding an endpoint does not change the probability.

Answer

1. \(Y\) is discrete; \(Z\) is continuous. 2. \(P(Y\le 3)=\frac{1}{12}\approx 0.0833\) 3. \(P(6.5\le Z\le 7.5)\approx 0.1650\) 4. Endpoints can change a discrete probability because \(P(X=k)\) may be positive; they do not change a continuous probability because \(P(X=k)=0\).
52512612
Classify each random variable as discrete or continuous. Also state whether its set of possible values is finite, countably infinite, or uncountably infinite. a) The angle, in degrees, at which an ideal spinner with an infinitely thin pointer stops. b) The amount of money in a piggy bank with a fixed capacity, recorded to the nearest cent. c) The exact lifetime of a radioactive particle before it decays. d) The number of daylight hours at a location, rounded to one decimal place.

Hints

- Between two possible values, can you always find more possible values? - Distinguish the physical quantity from the way it is recorded or rounded. - A countably infinite set can be listed in a sequence like the natural numbers. - Intervals of real numbers are uncountably infinite.

Solution

1. a) The angle is continuous because any real value in \([0,360)\) is possible. The set of values is uncountably infinite. 2. b) The recorded amount is discrete because values change in increments of \(\$0.01\). The fixed capacity makes the set of possible values finite. 3. c) The lifetime is continuous because it can take any positive real value. The set of values is uncountably infinite. 4. d) The rounded daylight hours are discrete because only values on a tenth-hour grid within the physical range from \(0.0\) through \(24.0\) can be recorded. The set is finite.

Answer

a) continuous; uncountably infinite b) discrete; finite c) continuous; uncountably infinite d) discrete; finite
52537312
A discrete random variable \(X\) has a binomial distribution with \(n=100\) and \(p=0.5\). A continuous random variable \(Y\) is normally distributed with \(\mu=50\) and \(\sigma=5\). 1. Find \(P(45\le X\le 55)\) and \(P(45<X<55)\). 2. Find \(P(45\le Y\le 55)\) using the empirical rule or a standard normal table. 3. Explain the key difference between discrete and continuous random variables by comparing probabilities such as \(P(X=45)\) and \(P(Y=45)\).

Hints

- Use differences of cumulative binomial probabilities for each interval. - How does removing the endpoints affect a sum compared with an integral? - Recall that continuous probabilities are areas under density curves. - What area is contributed by one vertical line above a fixed x-value?

Solution

1. For the binomial random variable, \(P(45\le X\le 55)=P(X\le 55)-P(X\le 44)\approx 0.7287\). Also, \(P(45<X<55)=P(X\le 54)-P(X\le 45)\approx 0.6318\). The probabilities differ because the endpoints \(45\) and \(55\) have positive probability. 2. The interval \([45,55]\) is \([\mu-\sigma,\mu+\sigma]\). By the empirical rule, \(P(45\le Y\le 55)\approx 0.6827\). 3. A discrete random variable may assign positive probability to an individual value; here, \(P(X=45)\approx 0.0485\). For a continuous random variable, \(P(Y=45)=0\). Therefore, including or excluding endpoints changes a discrete probability but not a continuous probability.

Answer

1. \(P(45\le X\le 55)\approx 0.7287\) and \(P(45<X<55)\approx 0.6318\) 2. \(P(45\le Y\le 55)\approx 0.6827\), or about \(68.3\%\) 3. A discrete variable can have \(P(X=k)>0\), so endpoints can matter. A continuous variable has \(P(Y=k)=0\), so endpoints do not affect interval probabilities.
53096212
A contest has three problems. Each problem earns \(0\) points for an incorrect solution, \(3\) points for a partially correct solution, or \(10\) points for a fully correct solution. The random variable \(X\) is a contestant’s total score on the three problems. Give the set of all possible values of \(X\).

Hints

- Start with the possible scores on one problem. - Organize the cases by the number of \(10\)-point scores. - List all totals formed from three allowed scores. - Include each total only once in the final set.

Solution

1. The possible score on each problem is in \(\{0, 3, 10\}\). 2. With no \(10\)-point scores, the possible totals are \(0\), \(3\), \(6\), and \(9\). 3. With exactly one \(10\)-point score, the possible totals are \(10\), \(13\), and \(16\). 4. With exactly two \(10\)-point scores, the possible totals are \(20\) and \(23\). 5. With three \(10\)-point scores, the total is \(30\). 6. Combining the distinct totals gives \(\{0, 3, 6, 9, 10, 13, 16, 20, 23, 30\}\).

Answer

\(\{0, 3, 6, 9, 10, 13, 16, 20, 23, 30\}\)
53097212
An urn contains five balls labeled \(1\), \(2\), \(3\), \(4\), and \(5\). Two balls are drawn with replacement. Define the random variable \(Z\) as follows: - \(Z=10\) if the sum of the two numbers is greater than \(7\). - \(Z=5\) if the sum is exactly \(7\). - \(Z=-5\) if the sum is less than \(7\). Determine the probability distribution of \(Z\).

Hints

- Count all ordered pairs from two draws with replacement. - List the pairs whose sum is exactly \(7\). - Use a \(5 \times 5\) grid or another systematic list to avoid missing pairs. - Check that the three probabilities add to \(1\).

Solution

1. There are \(5\cdot5=25\) equally likely ordered pairs. 2. A sum of \(7\) occurs for \((2, 5)\), \((3, 4)\), \((4, 3)\), and \((5, 2)\), so \(P(Z=5)=\frac{4}{25}=0.16\). 3. A sum less than \(7\) occurs for \(15\) ordered pairs, so \(P(Z=-5)=\frac{15}{25}=0.60\). 4. A sum greater than \(7\) occurs for the remaining \(6\) ordered pairs, so \(P(Z=10)=\frac{6}{25}=0.24\).

Answer

<table border="1"> <tr> <td>\(z\)</td> <td>\(-5\)</td> <td>\(5\)</td> <td>\(10\)</td> </tr> <tr> <td>\(P(Z=z)\)</td> <td>\(0.60\)</td> <td>\(0.16\)</td> <td>\(0.24\)</td> </tr> </table>
53097812
A delivery company compares the annual mileage of two vehicle types. The random variables \(X_1\) and \(X_2\) represent annual mileage, in thousands of miles, for a delivery van and a heavy truck, respectively. The table gives cumulative relative frequencies. <table> <thead> <tr> <th>Annual mileage up to ... thousand miles</th> <th>Delivery van \(X_1\)</th> <th>Heavy truck \(X_2\)</th> </tr> </thead> <tbody> <tr><td>\(10\)</td><td>\(0.12\)</td><td>\(0.04\)</td></tr> <tr><td>\(20\)</td><td>\(0.38\)</td><td>\(0.15\)</td></tr> <tr><td>\(30\)</td><td>\(0.65\)</td><td>\(0.32\)</td></tr> <tr><td>\(40\)</td><td>\(0.84\)</td><td>\(0.58\)</td></tr> <tr><td>\(50\)</td><td>\(0.95\)</td><td>\(0.82\)</td></tr> <tr><td>\(60\)</td><td>\(1.00\)</td><td>\(1.00\)</td></tr> </tbody> </table> a) Find the probability that a randomly selected heavy truck travels more than \(40{,}000\) miles in a year. b) Compare \(P(20<X_1\le40)\) and \(P(20<X_2\le40)\). Which vehicle type is more likely to have annual mileage in this interval? c) For delivery vans, which interval has the greatest probability: \([0, 20]\), \((20, 40]\), or \((40, 60]\)?

Hints

- Each table entry already includes all observations up to that mileage. - Find an interval probability by subtracting cumulative values at its endpoints. - Read the two vehicle columns separately. - Compute the probability in each proposed interval before comparing them.

Solution

1. \(P(X_2>40)=1-P(X_2\le40)=1-0.58=0.42\). 2. For delivery vans, \(P(20<X_1\le40)=0.84-0.38=0.46\). For heavy trucks, \(P(20<X_2\le40)=0.58-0.15=0.43\). The delivery-van probability is greater. 3. For delivery vans, the three interval probabilities are \(0.38\), \(0.84-0.38=0.46\), and \(1.00-0.84=0.16\). Therefore, \((20, 40]\) has the greatest probability.

Answer

a) \(P(X_2>40)=0.42\) b) \(P(20<X_1\le40)=0.46\) and \(P(20<X_2\le40)=0.43\); delivery vans have the greater probability. c) The interval \((20, 40]\) has the greatest probability, \(0.46\).
53115612
Two fair four-sided dice labeled \(1\) through \(4\) are rolled. The random variable \(Y\) is the sum of the two results. 1. Determine the set of possible values of \(Y\). 2. Calculate the probability distribution of \(Y\) and present it in a table. 3. Use this example to explain the difference between an outcome of the experiment and a value of the random variable.

Hints

- Organize the \(16\) ordered outcomes in a grid. - Count how many outcomes produce each possible sum. - Decide whether a sum such as \(5\) is itself an outcome or a quantity calculated from an outcome. - Think of a random variable as a function from outcomes to numbers.

Solution

1. The smallest sum is \(1+1=2\), and the largest is \(4+4=8\). Thus, the possible values are \(\{2,3,4,5,6,7,8\}\). 2. There are \(4\cdot4=16\) equally likely ordered outcomes. The numbers of outcomes producing sums \(2\) through \(8\) are \(1,2,3,4,3,2,1\), respectively. Dividing by \(16\) gives the distribution. 3. An outcome is an ordered pair such as \((1, 3)\), describing the result on each die. The random variable maps that outcome to the value \(Y=4\). Several different outcomes can produce the same value of \(Y\).

Answer

1. \(\{2,3,4,5,6,7,8\}\) 2. <table> <tr><td>\(k\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr> <tr><td>\(P(Y=k)\)</td><td>\(\frac{1}{16}\)</td><td>\(\frac{2}{16}\)</td><td>\(\frac{3}{16}\)</td><td>\(\frac{4}{16}\)</td><td>\(\frac{3}{16}\)</td><td>\(\frac{2}{16}\)</td><td>\(\frac{1}{16}\)</td></tr> </table> 3. An outcome is a specific ordered pair, while a value of \(Y\) is the sum assigned to that outcome. Different outcomes may have the same sum.
53118312
A fair six-sided die and a fair eight-sided die, each labeled consecutively starting at \(1\), are rolled. The random variable \(X\) is the sum of the two results. a) Determine the probability distribution of \(X\) in a table. b) Find the probability that the sum is a perfect square. c) Find \(P(5<X\le9)\).

Hints

- Count the total ordered outcomes. - Use a grid to tally each possible sum. - Identify the perfect squares between \(2\) and \(14\). - Translate the compound inequality carefully.

Solution

1. There are \(6\cdot8=48\) equally likely ordered outcomes, and the possible sums are \(2\) through \(14\). 2. The numbers of outcomes producing these sums are \(1,2,3,4,5,6,6,6,5,4,3,2,1\), respectively. Divide each count by \(48\) to obtain the distribution. 3. The perfect-square sums are \(4\) and \(9\). Thus, the probability is \(\frac{3}{48}+\frac{6}{48}=\frac{3}{16}=0.1875\). 4. The sums satisfying \(5<X\le9\) are \(6\), \(7\), \(8\), and \(9\). Their probability is \(\frac{5+6+6+6}{48}=\frac{23}{48}\approx0.4792\).

Answer

a) <table> <tr><td>\(x\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(\frac{1}{48}\)</td><td>\(\frac{2}{48}\)</td><td>\(\frac{3}{48}\)</td><td>\(\frac{4}{48}\)</td><td>\(\frac{5}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{5}{48}\)</td><td>\(\frac{4}{48}\)</td><td>\(\frac{3}{48}\)</td><td>\(\frac{2}{48}\)</td><td>\(\frac{1}{48}\)</td></tr> </table> b) \(P(\text{perfect square})=\frac{3}{16}=0.1875\) c) \(P(5<X\le9)=\frac{23}{48}\approx0.4792\)
53602112
The spinner shown has \(6\) equal sections. It is spun twice. The random variable \(Y\) is the product of the two numbers. Create a probability distribution table for \(Y\).
Figure for problem 536021

Hints

- Determine the probability of each number on one spin. - List the different products that two spins can produce. - A tree diagram can help organize all ordered pairs. - Several different ordered pairs produce a product of \(0\). - Check that the probabilities add to \(1\).

Solution

1. On one spin, \(P(0)=\frac{1}{6}\), \(P(1)=\frac{2}{6}=\frac{1}{3}\), and \(P(2)=\frac{3}{6}=\frac{1}{2}\). 2. The possible products are \(0\), \(1\), \(2\), and \(4\). 3. The product is \(0\) when at least one spin is \(0\), so \(P(Y=0)=1-\left(\frac{5}{6}\right)^2=\frac{11}{36}\). 4. \(P(Y=1)=\left(\frac{1}{3}\right)^2=\frac{1}{9}\). 5. \(P(Y=2)=2\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{3}\). 6. \(P(Y=4)=\left(\frac{1}{2}\right)^2=\frac{1}{4}\). 7. The probabilities add to \(1\).

Answer

<table> <tr><td>\(y\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td></tr> <tr><td>\(P(Y=y)\)</td><td>\(\frac{11}{36}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{4}\)</td></tr> </table>
54725812
A discrete random variable has distribution <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><td>\(P(X=x)\)</td><td>\(0.20\)</td><td>\(0.30\)</td><td>\(0.30\)</td><td>\(0.20\)</td></tr> </table> An integer \(m\) is a median if \(P(X\le m)\ge0.5\) and \(P(X\ge m)\ge0.5\). Find every integer median and explain why a discrete distribution can have more than one.

Hints

- Check both inequalities in the median definition for each candidate. - Build cumulative probability from the left and from the right. - Discrete probability can put exactly half the mass on either side of a gap.

Solution

1. For \(m=1\), \(P(X\le1)=0.50\) and \(P(X\ge1)=0.80\), so \(1\) is a median. 2. For \(m=2\), \(P(X\le2)=0.80\) and \(P(X\ge2)=0.50\), so \(2\) is a median. 3. For \(m=0\), the lower cumulative probability is only \(0.20\); for \(m=3\), the upper probability is only \(0.20\). 4. Therefore the integer medians are \(1\) and \(2\). A discrete distribution can have more than one median because two adjacent values can both leave at least half the probability at or below and at or above the candidate value.

Answer

The integer medians are \(1\) and \(2\). A discrete distribution can have more than one median because multiple values can satisfy both cumulative-probability inequalities.
54728712
Let \(I_A,I_B,I_C\) be the indicator variables for three events \(A,B,C\). a) Prove that if the events are pairwise mutually exclusive, then \(I_{A\cup B\cup C}=I_A+I_B+I_C\) for every outcome. b) Give an outcome-level explanation of why the identity fails when an outcome belongs to both \(A\) and \(B\).

Hints

- Indicators translate set membership into zeros and ones. - Check separately outcomes in none of the events and outcomes in one event. - Overlap affects a sum of indicators differently from an indicator of a union.

Solution

1. For pairwise mutually exclusive events, an outcome belongs to at most one of \(A,B,C\). 2. If it belongs to none, both sides equal \(0\). If it belongs to exactly one, the union indicator is \(1\) and exactly one indicator on the right is \(1\), so the sum is also \(1\). 3. If an outcome belongs to both \(A\) and \(B\), the union indicator is still \(1\), but \(I_A+I_B+I_C\ge2\). Thus overlap causes double-counting.

Answer

a) The identity holds pointwise because at most one indicator can equal \(1\). b) On an outcome in \(A\cap B\), the left side is \(1\) while the right side is at least \(2\).
54733412
A discrete random variable \(X\) has the distribution \(P(X=0)=0.10\), \(P(X=1)=0.20\), \(P(X=2)=0.30\), and \(P(X=3)=0.40\). Condition on the event \(X\ge1\). a) Find the conditional distribution of \(X\). b) Find \(P(X\ge2\mid X\ge1)\).

Hints

- Add the probabilities of all values retained by the condition. - Divide each retained original probability by that conditioning probability. - For part b, combine the retained probabilities for \(X=2\) and \(X=3\).

Solution

1. The conditioning event has probability \(P(X\ge1)=0.20+0.30+0.40=0.90\). 2. Renormalizing the retained values gives \(P(X=1\mid X\ge1)=\frac{0.20}{0.90}=\frac29\), \(P(X=2\mid X\ge1)=\frac{0.30}{0.90}=\frac13\), and \(P(X=3\mid X\ge1)=\frac{0.40}{0.90}=\frac49\). The conditional probability at \(X=0\) is \(0\). 3. \(P(X\ge2\mid X\ge1)=\frac{0.30+0.40}{0.90}=\frac79\).

Answer

a) \(P(X=0\mid X\ge1)=0\), \(P(X=1\mid X\ge1)=\frac29\), \(P(X=2\mid X\ge1)=\frac13\), and \(P(X=3\mid X\ge1)=\frac49\). b) \(P(X\ge2\mid X\ge1)=\frac79\).
54739712
Five data logs are placed in a random order. Two logs are flagged and three are unflagged. An analyst reviews the logs in order and stops at the first flagged log. Let \(X\) be the number of unflagged logs reviewed before stopping. a) Construct the probability distribution of \(X\). b) Find \(P(X\ge 2)\).

Hints

- List how many unflagged logs could appear before the first flagged log. - For each possible value, follow the sequence of log types that must occur. - Remember that the composition of the remaining logs changes after each review.

Solution

1. The possible values are \(0,1,2,3\). 2. The first log is flagged with probability \(\frac{2}{5}\), so \(P(X=0)=\frac{2}{5}\). 3. Continuing without replacement gives \(P(X=1)=\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\), \(P(X=2)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}\), and \(P(X=3)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{10}\). 4. Therefore, \(P(X\ge 2)=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{2}{5}\)</td><td>\(\frac{3}{10}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{1}{10}\)</td></tr> </table> b) \(P(X\ge 2)=\frac{3}{10}=0.30\)
54739912
Three name cards labeled \(A\), \(B\), and \(C\) are randomly placed into three slots labeled \(A\), \(B\), and \(C\), one card per slot. Let \(X\) be the number of cards placed in the slot with the same label. A student claims that \(X\) can be \(0,1,2,\) or \(3\), with probability \(\frac{1}{4}\) for each value. Identify the errors in the claim and give the correct probability distribution of \(X\).

Hints

- List the possible placements rather than assuming the values of the random variable are equally likely. - Check whether exactly two matching positions can occur without forcing the third match. - Count placements that produce each attainable value.

Solution

1. There are \(3!=6\) equally likely placements. 2. Two placements have no matches, three placements have exactly one match, and one placement has three matches. 3. Exactly two matches are impossible: if two cards are in their matching slots, the remaining card must also be in its matching slot. 4. Thus, \(P(X=0)=\frac{2}{6}=\frac{1}{3}\), \(P(X=1)=\frac{3}{6}=\frac{1}{2}\), and \(P(X=3)=\frac{1}{6}\).

Answer

The claim has two errors: \(X=2\) is impossible, and the possible values are not equally likely. <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{6}\)</td></tr> </table>
54740012
A discrete random variable \(X\) has the probability distribution shown. <table> <tr><th>\(x\)</th><th>\(-2\)</th><th>\(-1\)</th><th>\(0\)</th><th>\(1\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{10}\)</td><td>\(\frac{1}{5}\)</td><td>\(\frac{7}{20}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{10}\)</td></tr> </table> Define \(Y=|X-1|\). a) Construct the probability distribution of \(Y\). b) Find \(P(Y\le 1)\).

Hints

- Apply the definition of the new random variable to every possible value of the original one. - Different original values may lead to the same transformed value. - Check that the probabilities in the new distribution still total one.

Solution

1. Transform each possible value of \(X\): \(-2\mapsto3\), \(-1\mapsto2\), \(0\mapsto1\), \(1\mapsto0\), and \(3\mapsto2\). 2. Values of \(X\) that produce the same value of \(Y\) must have their probabilities combined. 3. Therefore, \(P(Y=0)=\frac{1}{4}\), \(P(Y=1)=\frac{7}{20}\), \(P(Y=2)=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}\), and \(P(Y=3)=\frac{1}{10}\). 4. Thus, \(P(Y\le1)=\frac{1}{4}+\frac{7}{20}=\frac{3}{5}\).

Answer

a) <table> <tr><th>\(y\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(Y=y)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{7}{20}\)</td><td>\(\frac{3}{10}\)</td><td>\(\frac{1}{10}\)</td></tr> </table> b) \(P(Y\le1)=\frac{3}{5}=0.60\)
54740312
A fair coin is tossed three times. Let \(X\) be the length of the longest consecutive run of heads. Define \(X=0\) when no heads occur. a) Construct the probability distribution of \(X\). b) Find \(P(X\ge2)\).

Hints

- List all equally likely toss sequences before classifying them. - For each sequence, look for the longest uninterrupted block of heads. - Several different sequences can produce the same value of the random variable.

Solution

1. The eight equally likely sequences are classified by their longest run of heads. 2. The sequence \(TTT\) gives \(X=0\); four sequences give \(X=1\); two sequences give \(X=2\); and \(HHH\) gives \(X=3\). 3. Thus, the probabilities are \(\frac{1}{8},\frac{4}{8},\frac{2}{8},\frac{1}{8}\) for \(X=0,1,2,3\), respectively. 4. Therefore, \(P(X\ge2)=\frac{2}{8}+\frac{1}{8}=\frac{3}{8}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{8}\)</td><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{8}\)</td></tr> </table> b) \(P(X\ge2)=\frac{3}{8}\)
54740412
A discrete random variable \(X\) can take only the values \(-2,0,1,4\). You are given \(P(X<0)=0.18\), \(P(X\le0)=0.47\), and \(P(X>1)=0.22\). a) Determine the probability distribution of \(X\). b) Find \(P(|X|\le1)\).

Hints

- Match each stated event to the possible values it contains. - Subtract nested event probabilities to isolate a single value. - Use the fact that all point probabilities must add to one.

Solution

1. Since \(-2\) is the only negative value, \(P(X=-2)=0.18\). 2. The difference between \(P(X\le0)\) and \(P(X<0)\) gives \(P(X=0)=0.47-0.18=0.29\). 3. Since \(4\) is the only value greater than \(1\), \(P(X=4)=0.22\). 4. The remaining probability is \(P(X=1)=1-0.18-0.29-0.22=0.31\). 5. The event \(|X|\le1\) contains \(X=0\) and \(X=1\), so its probability is \(0.29+0.31=0.60\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(-2\)</th><th>\(0\)</th><th>\(1\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.18\)</td><td>\(0.29\)</td><td>\(0.31\)</td><td>\(0.22\)</td></tr> </table> b) \(P(|X|\le1)=0.60\)
54740512
A testing system randomly selects one of two calibration modules. Module \(A\) is selected with probability \(0.70\), and module \(B\) is selected with probability \(0.30\). Module \(A\) reports \(0\) or \(1\) with equal probability. Module \(B\) reports \(1\) with probability \(0.20\) and \(2\) with probability \(0.80\). Let \(X\) be the reported value. a) Construct the probability distribution of \(X\). b) Find \(P(X\ne1)\).

Hints

- Separate the selection of a module from the value reported by that module. - Check whether each reported value can arise through one route or through both routes. - Add probabilities only after accounting for every route to the same value.

Solution

1. A report of \(0\) can occur only through module \(A\), so \(P(X=0)=0.70\cdot0.50=0.35\). 2. A report of \(1\) can occur through either module, so \(P(X=1)=0.70\cdot0.50+0.30\cdot0.20=0.41\). 3. A report of \(2\) can occur only through module \(B\), so \(P(X=2)=0.30\cdot0.80=0.24\). 4. Therefore, \(P(X\ne1)=0.35+0.24=0.59\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.35\)</td><td>\(0.41\)</td><td>\(0.24\)</td></tr> </table> b) \(P(X\ne1)=0.59\)
54740712
For a real number \(a\), the table is intended to define the probability distribution of a discrete random variable \(X\). <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(a\)</td><td>\(2a\)</td><td>\(\frac{1}{2}-a\)</td><td>\(\frac{1}{2}-2a\)</td></tr> </table> Find all values of \(a\) for which the table is a valid probability distribution and \(P(X\ge2)>P(X\le1)\).

Hints

- Check both requirements for a probability distribution, not only the total. - Express each event probability by adding the relevant table entries. - Intersect the restrictions rather than treating them separately.

Solution

1. The four expressions always sum to \(1\). 2. Nonnegative probabilities require \(a\ge0\), \(\frac{1}{2}-a\ge0\), and \(\frac{1}{2}-2a\ge0\), giving \(0\le a\le\frac{1}{4}\). 3. The event probabilities are \(P(X\ge2)=1-3a\) and \(P(X\le1)=3a\). 4. The required comparison gives \(1-3a>3a\), so \(a<\frac{1}{6}\). 5. Combining the conditions gives \(0\le a<\frac{1}{6}\).

Answer

\(0\le a<\frac{1}{6}\)
54740912
A wildlife camera records one of four animal categories on each of three independent nights. Each category is equally likely on each night. Let \(X\) be the number of distinct categories recorded during the three nights. a) Construct the probability distribution of \(X\). b) Find the probability that exactly one category repeats and a second category appears once.

Hints

- Count ordered three-night records, not unordered sets. - The all-same and all-different cases are easiest to count directly. - Interpret the wording in part b in terms of the number of distinct categories.

Solution

1. There are \(4^3=64\) equally likely category sequences. 2. Exactly one distinct category occurs in \(4\) sequences, so \(P(X=1)=\frac{4}{64}=\frac{1}{16}\). 3. Exactly three distinct categories occur in \(4\cdot3\cdot2=24\) sequences, so \(P(X=3)=\frac{24}{64}=\frac{3}{8}\). 4. The remaining \(64-4-24=36\) sequences use exactly two distinct categories, so \(P(X=2)=\frac{36}{64}=\frac{9}{16}\). 5. The description in part b is exactly the event \(X=2\), so the probability is \(\frac{9}{16}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{16}\)</td><td>\(\frac{9}{16}\)</td><td>\(\frac{3}{8}\)</td></tr> </table> b) \(\frac{9}{16}\)
54741412
An integer \(N\) is selected uniformly at random from \(1\) through \(12\). Define \(X\) as the remainder when \(N\) is divided by \(4\), and define \(Y=\left\lfloor\frac{N-1}{3}\right\rfloor\). a) Construct the probability distributions of \(X\) and \(Y\). b) Explain why \(X\) and \(Y\) have the same distribution even though they are not the same random variable.

Hints

- Evaluate each definition across the twelve equally likely integers. - Compare frequencies of output values, not which inputs produce them. - To show the variables differ, look for one input where their outputs are unequal.

Solution

1. Each remainder \(0,1,2,3\) occurs for exactly \(3\) of the \(12\) integers, so \(P(X=x)=\frac{1}{4}\) for each value. 2. The values \(N=1,2,3\) give \(Y=0\); \(4,5,6\) give \(Y=1\); \(7,8,9\) give \(Y=2\); and \(10,11,12\) give \(Y=3\). Thus, \(P(Y=y)=\frac{1}{4}\) for each value. 3. The distributions match because both variables assign probability \(\frac{1}{4}\) to each of \(0,1,2,3\). 4. They are not the same random variable because, for example, \(N=1\) gives \(X=1\) but \(Y=0\).

Answer

a) Both variables have the distribution <table> <tr><th>Value</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{4}\)</td></tr> </table> b) They assign the same probabilities to the same possible values, but they can give different values for the same selected integer; for example, \(N=1\) gives \(X=1\) and \(Y=0\).
54741512
A discrete random variable \(X\) has the distribution shown. <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.10\)</td><td>\(0.20\)</td><td>\(0.30\)</td><td>\(0.25\)</td><td>\(0.15\)</td></tr> </table> After learning that \(X\ge2\), construct the conditional probability distribution of \(X\). Then find the conditional probability that \(X\) is odd.

Hints

- Remove values that are impossible after the new information is known. - Rescale the remaining probabilities so that they total one. - Identify the odd values within the restricted set.

Solution

1. The conditioning event has probability \(P(X\ge2)=0.30+0.25+0.15=0.70\). 2. Divide each remaining point probability by \(0.70\): \(P(X=2\mid X\ge2)=\frac{3}{7}\), \(P(X=3\mid X\ge2)=\frac{5}{14}\), and \(P(X=4\mid X\ge2)=\frac{3}{14}\). 3. Within the conditional distribution, the only odd value is \(3\), so \(P(X\text{ is odd}\mid X\ge2)=\frac{5}{14}\).

Answer

<table> <tr><th>\(x\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(X=x\mid X\ge2)\)</th><td>\(\frac{3}{7}\)</td><td>\(\frac{5}{14}\)</td><td>\(\frac{3}{14}\)</td></tr> </table> \(P(X\text{ is odd}\mid X\ge2)=\frac{5}{14}\)
54741612
The cumulative distribution function of a discrete random variable \(X\) is \(F(x)=0\) for \(x<0\), \(F(x)=0.20\) for \(0\le x<2\), \(F(x)=0.65\) for \(2\le x<5\), and \(F(x)=1\) for \(x\ge5\). a) Construct the probability distribution of \(X\). b) Find \(P(0<X\le5)\).

Hints

- Look for the points where the cumulative function changes value. - Each point probability equals the size of the jump at that point. - Apply both endpoint conditions when selecting values for the event.

Solution

1. The possible values occur where the cumulative distribution jumps: \(0,2,5\). 2. The jump at \(0\) is \(0.20\), so \(P(X=0)=0.20\). 3. The jump at \(2\) is \(0.65-0.20=0.45\), and the jump at \(5\) is \(1-0.65=0.35\). 4. The event \(0<X\le5\) contains \(X=2\) and \(X=5\), so its probability is \(0.45+0.35=0.80\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(2\)</th><th>\(5\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(0.20\)</td><td>\(0.45\)</td><td>\(0.35\)</td></tr> </table> b) \(P(0<X\le5)=0.80\)
54741912
A fair four-sided die has faces labeled \(0,1,1,3\). It is rolled twice independently. Let \(X\) be the larger of the two displayed numbers. a) Construct the probability distribution of \(X\). b) Find \(P(X<3)\).

Hints

- Consider cumulative events such as both rolls being at most a certain value. - For the middle value, remove the case where the maximum is smaller. - Use the total probability to obtain the final point probability.

Solution

1. A single roll is \(0\) with probability \(\frac{1}{4}\), at most \(1\) with probability \(\frac{3}{4}\), and \(3\) with probability \(\frac{1}{4}\). 2. Both rolls must be \(0\) for \(X=0\), so \(P(X=0)=\left(\frac{1}{4}\right)^2=\frac{1}{16}\). 3. For \(X=1\), both rolls must be at most \(1\), but not both \(0\): \(P(X=1)=\left(\frac{3}{4}\right)^2-\frac{1}{16}=\frac{1}{2}\). 4. The remaining probability is \(P(X=3)=1-\frac{1}{16}-\frac{1}{2}=\frac{7}{16}\). 5. Therefore, \(P(X<3)=\frac{1}{16}+\frac{1}{2}=\frac{9}{16}\).

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{16}\)</td><td>\(\frac{1}{2}\)</td><td>\(\frac{7}{16}\)</td></tr> </table> b) \(P(X<3)=\frac{9}{16}\)
54742112
Four debaters \(A,B,C,D\) are divided uniformly at random into two unlabeled teams of two. Before the random division, \(A\) and \(B\) are partners, and \(C\) and \(D\) are partners. Let \(X\) be the number of original partner pairs that remain together. a) Construct the probability distribution of \(X\). b) Explain why \(X=1\) is impossible.

Hints

- List the distinct ways to partition four people into two pairs. - Count retained original pairs for each partition. - Consider what happens to the two remaining people if one original pair stays together.

Solution

1. The three equally likely team divisions are \(AB/CD\), \(AC/BD\), and \(AD/BC\). 2. The first division keeps both original pairs together, so \(X=2\). Each other division separates both original pairs, so \(X=0\). 3. Therefore, \(P(X=0)=\frac{2}{3}\) and \(P(X=2)=\frac{1}{3}\). 4. If one original pair remains together, the two remaining debaters must form the other team, so the second original pair also remains together. Hence \(X=1\) cannot occur.

Answer

a) <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{2}{3}\)</td><td>\(\frac{1}{3}\)</td></tr> </table> b) Keeping either original pair together forces the other two debaters, who are also original partners, to form the second team. Therefore, exactly one retained pair is impossible.
54742212
A discrete random variable \(X\) has possible values \(1,3,k\), where \(k\) is an integer greater than \(3\). The corresponding probabilities are \(0.20,0.50,0.30\). You also know that \(P(X\le5)=0.70\) and \(P(X<8)=1\). Determine all possible values of \(k\).

Hints

- Identify which known values already account for the stated cumulative probability. - Use each event statement to place an upper or lower bound on the unknown support value. - Apply the requirement that the unknown is an integer.

Solution

1. Since \(P(X\le5)=0.70\) equals the combined probability at \(1\) and \(3\), the value \(k\) must be greater than \(5\). 2. Since \(P(X<8)=1\), every possible value must be less than \(8\), so \(k<8\). 3. The integer values satisfying \(5<k<8\) are \(k=6\) and \(k=7\).

Answer

\(k\in\{6,7\}\)
54742812
For a selected station in the diagram, let \(X\) be the number of track segments incident to that station. Method A selects one station uniformly at random. Method B selects one track segment uniformly at random and then selects one of that segment's two endpoint stations uniformly at random. a) Construct the probability distribution of \(X\) under each method. b) Explain why the two methods give different distributions.
Figure for problem 547428

Hints

- Identify the station types and the value of the random variable for each type. - Determine the equally likely elementary outcomes separately for the two methods. - Under the second method, count how often each station can be reached through a selected segment.

Solution

1. The diagram has \(5\) stations. Under Method A, two endpoint stations have \(X=1\) and three interior stations have \(X=2\). Thus, the probabilities are \(\frac{2}{5}\) and \(\frac{3}{5}\). 2. Under Method B, there are \(4\cdot2=8\) equally likely segment-endpoint selections. The two endpoint stations each appear once, while the three interior stations each appear twice. 3. Therefore, Method B gives \(P(X=1)=\frac{2}{8}=\frac{1}{4}\) and \(P(X=2)=\frac{6}{8}=\frac{3}{4}\). 4. The methods differ because an interior station belongs to two segments and is therefore twice as likely as an endpoint under Method B, while all stations are equally likely under Method A.

Answer

a) Method A: <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{2}{5}\)</td><td>\(\frac{3}{5}\)</td></tr> </table> Method B: <table> <tr><th>\(x\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{3}{4}\)</td></tr> </table> b) Method B favors interior stations because each is an endpoint of two track segments, while each end station belongs to only one segment.
52575012
An urn contains three balls labeled \(1\), \(2\), and \(3\). A ball is drawn three times with replacement. Let the results be \(z_1\), \(z_2\), and \(z_3\), and define the partial sums \(s_1=z_1\), \(s_2=z_1+z_2\), and \(s_3=z_1+z_2+z_3\). The random variable \(X\) is the number of these three partial sums that are even. Determine the probability distribution of \(X\) and present it in a table.

Hints

- Build a tree that records only whether each draw is odd or even. - At each endpoint, determine the parity of \(s_1\), \(s_2\), and \(s_3\), then count the even partial sums. - The probabilities of an even draw and an odd draw are not equal. - Check that all probabilities in the distribution add to \(1\).

Solution

1. A draw is even with probability \(\frac{1}{3}\) and odd with probability \(\frac{2}{3}\). Let \(E\) and \(O\) represent the parity of each draw. 2. Track the parity of the partial sums for each draw-parity pattern. The pattern \(OEE\) gives no even partial sums, so \(P(X=0)=\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}=\frac{2}{27}\). 3. The patterns \(OEO\), \(OOO\), and \(EOE\) each give exactly one even partial sum. Their total probability is \(\frac{4}{27}+\frac{8}{27}+\frac{2}{27}=\frac{14}{27}\). 4. The patterns \(OOE\), \(EOO\), and \(EEO\) each give exactly two even partial sums. Their total probability is \(\frac{4}{27}+\frac{4}{27}+\frac{2}{27}=\frac{10}{27}\). 5. The pattern \(EEE\) gives three even partial sums, so \(P(X=3)=\frac{1}{27}\). 6. The probabilities add to \(\frac{2+14+10+1}{27}=1\).

Answer

<table border="1"> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(\frac{2}{27}\)</td> <td>\(\frac{14}{27}\)</td> <td>\(\frac{10}{27}\)</td> <td>\(\frac{1}{27}\)</td> </tr> </table>
53097612
In a table tennis match between Maya and Noah, Maya wins each game with probability \(0.6\). The match is best-of-five, so the first player to win \(3\) games wins the match. a) Determine the probability distribution of the random variable \(X\), the total number of games played. b) Find the probability that Maya wins the match.

Hints

- The two players do not have equal game-winning probabilities. - To win in exactly \(4\) games, the winner must win game \(4\) and exactly \(2\) of the first \(3\) games. - Add Maya’s probabilities of winning in \(3\), \(4\), and \(5\) games. - Check that the probabilities of all possible match lengths add to \(1\).

Solution

1. The possible match lengths are \(3\), \(4\), and \(5\) games. Let \(p=0.6\) and \(q=0.4\). 2. A three-game match is a sweep by either player, so \(P(X=3)=p^3+q^3=0.216+0.064=0.28\). 3. A four-game match requires the winner to have exactly \(2\) wins in the first \(3\) games and then win game \(4\). Thus, \(P(X=4)=\binom{3}{2}p^3q+\binom{3}{2}q^3p=0.2592+0.1152=0.3744\). 4. A five-game match requires a \(2\)-to-\(2\) split after four games, so \(P(X=5)=\binom{4}{2}p^2q^2=0.3456\). 5. Maya can win in \(3\), \(4\), or \(5\) games. Her total win probability is \(p^3+\binom{3}{2}p^3q+\binom{4}{2}p^3q^2=0.68256\).

Answer

a) \(P(X=3)=0.28\), \(P(X=4)=0.3744\), and \(P(X=5)=0.3456\). b) Maya wins the match with probability \(0.68256\approx68.3\%\).
54726712
Random variables \(X\) and \(Y\) take values \(0,2,4\). Their probability distributions are <table> <tr><th></th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th></tr> <tr><td>\(X\)</td><td>\(0.40\)</td><td>\(0.40\)</td><td>\(0.20\)</td></tr> <tr><td>\(Y\)</td><td>\(0.20\)</td><td>\(0.40\)</td><td>\(0.40\)</td></tr> </table> Compare \(P(X\ge t)\) and \(P(Y\ge t)\) at every threshold where the comparison can change. Determine whether one variable has at least as large a tail probability at every threshold, and compare their expected values.

Hints

- Tail probabilities change only when a threshold passes a possible value. - Compare the full set of tail probabilities, not only the largest outcome. - Use the probability-weighted values for the final mean comparison.

Solution

1. For thresholds above \(0\) and at most \(2\), \(P(X\ge t)=0.60\) and \(P(Y\ge t)=0.80\). 2. For thresholds above \(2\) and at most \(4\), \(P(X\ge t)=0.20\) and \(P(Y\ge t)=0.40\). 3. At thresholds at or below \(0\), both tail probabilities are \(1\); above \(4\), both are \(0\). 4. Thus \(P(Y\ge t)\ge P(X\ge t)\) for every \(t\), with strict inequality for some thresholds. Therefore \(Y\) has at least as large a probability of meeting or exceeding every threshold. 5. \(E(X)=0\cdot0.4+2\cdot0.4+4\cdot0.2=1.6\), and \(E(Y)=0\cdot0.2+2\cdot0.4+4\cdot0.4=2.4\).

Answer

For \(t\le0\), both tail probabilities are \(1\). For \(0<t\le2\), they are \(0.60\) for \(X\) and \(0.80\) for \(Y\). For \(2<t\le4\), they are \(0.20\) for \(X\) and \(0.40\) for \(Y\). For \(t>4\), both are \(0\). Therefore \(Y\) has at least as large a tail probability at every threshold. Their expected values are \(E(X)=1.6\) and \(E(Y)=2.4\).

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