Free math worksheets

- In an equally likely sample space, probability depends on event size. - Different subsets can have the same number of outcomes. - Compare membership, not only probability.

1. One event is \(\{1,2,3\}\). 2. Another is \(\{2,5,8\}\). 3. Each contains \(3\) of \(8\) equally likely outcomes, but the outcome sets differ.

For example, \(\{1,2,3\}\) and \(\{2,5,8\}\) each have probability \(\frac{3}{8}\), but they are different subsets of the sample space.

- An impossible event contains no outcomes. - A certain event contains every outcome. - Use mutually exclusive suit categories and exhaustive color categories.

1. “The card is both a heart and a spade” is impossible, so its probability is \(0\). 2. “The card is red or black” is certain, so its probability is \(1\). 3. These events correspond to the empty set and the full sample space.

Impossible: both a heart and a spade, probability \(0\). Certain: red or black, probability \(1\).

- Translate “at most” into an inequality. - Negate the entire inequality. - Use the fact that the count takes integer values.

1. “At most \(2\)” means \(X\le2\). 2. Its complement contains all outcomes with \(X>2\). 3. For integer-valued \(X\), this is \(X\ge3\).

The complement is “at least \(3\) goals,” written \(X\ge3\).

- What height must a rectangle of width \(40\) have so that its area is \(1\)? - Connect probability with area under the density curve. - What happens to area when an interval has width \(0\)? - Distinguish continuous measurements from discrete outcomes.

1. a) A uniform density on \([0,40]\) has height \(\frac{1}{40}\), so \(f(x)=\frac{1}{40}\) for \(0\le x\le 40\) and \(f(x)=0\) otherwise. 2. b) \(P(10\le X\le 30)=\int_{10}^{30}\frac{1}{40}\,dx=\left[\frac{x}{40}\right]_{10}^{30}=\frac{30-10}{40}=0.5\). 3. c) For a continuous random variable, \(P(X=k)=\int_k^k f(x)\,dx=0\). Both probabilities are \(0\), so the statement is mathematically correct.

a) \(f(x)=\frac{1}{40}\) for \(0\le x\le 40\), and \(f(x)=0\) otherwise. b) \(\int_{10}^{30}\frac{1}{40}\,dx=0.5\) c) The statement is true because both exact-value probabilities are \(0\).

- Use the difference of cumulative distribution values at the interval endpoints. - Interpret probability as area under a density curve. - What area is associated with an interval of width \(0\)?

1. \(P(2<Z\le 6)=F(6)-F(2)=0.80-0.35=0.45\). Since \(P(Z=2)=0\), \(P(2\le Z\le 6)=0.45\) as well. 2. For a continuous random variable, \(P(Z=a)=\int_a^a f(z)\,dz=0\) because the interval has zero width. 3. Including or excluding endpoints does not change the probability. Thus, \(P(a\le Z\le b)=P(a<Z\le b)=P(a\le Z<b)=P(a<Z<b)\).

1. \(P(2<Z\le 6)=0.45\) and \(P(2\le Z\le 6)=0.45\) 2. \(P(Z=a)=\int_a^a f(z)\,dz=0\) 3. Open, closed, and half-open intervals with the same endpoints have equal probabilities for a continuous random variable.

- List the possible color combinations when order does not matter. - A mixed-color pair can occur in two orders. - Compare the probabilities of the outcomes.

1. The possible unordered color outcomes are \(S=\{BB, BR, RR\}\), where \(BR\) means one ball of each color. 2. \(P(BB)=\frac{7}{10}\cdot\frac{7}{10}=0.49\). 3. The mixed-color outcome can occur in either order, so \(P(BR)=\frac{7}{10}\cdot\frac{3}{10}+\frac{3}{10}\cdot\frac{7}{10}=0.42\). 4. \(P(RR)=\frac{3}{10}\cdot\frac{3}{10}=0.09\). 5. The three outcomes have different probabilities, so this sample space is not an equally likely-outcomes model.

a) \(S=\{BB, BR, RR\}\), where \(BR\) means one ball of each color b) \(P(BB)=0.49\), \(P(BR)=0.42\), and \(P(RR)=0.09\) c) The outcomes are not equally likely because their probabilities are different.

- What must the probabilities of all six outcomes add to? - Let \(p\) represent the probability of each outcome from \(1\) through \(5\). - What assumption is required before counting favorable and possible outcomes gives a probability? - Compare a fair die with a loaded die.

1. Let \(p\) be the probability of rolling any one of the numbers \(1\) through \(5\). Then \(P(6)=2p\). Since all outcome probabilities add to \(1\), \(5p+2p=1\). Thus \(7p=1\), so \(p=\frac{1}{7}\) and \(P(6)=\frac{2}{7}\approx 0.2857\). 2. The favorable-outcomes-over-possible-outcomes formula assumes that all individual outcomes are equally likely. This die is loaded, so its six outcomes are not equally likely.

1. \(P(6)=\frac{2}{7}\approx 0.2857\) 2. The formula does not apply directly because the six outcomes are not equally likely.

- When can you assume that all outcomes have the same probability? - Which numbers from \(1\) through \(6\) are prime? - Use one variable for the probability of each outcome from \(1\) through \(5\). - Which probability rule determines the value of that variable?

1. For a fair die, the six outcomes are equally likely. The prime outcomes are \(2\), \(3\), and \(5\), so \(P(A)=\frac{3}{6}=\frac{1}{2}=0.5\). 2. Let \(x\) be the probability of each outcome from \(1\) through \(5\). Then \(P(6)=3x\). 3. The probabilities must add to \(1\): \(5x+3x=1\). Thus \(8x=1\), so \(x=0.125\). 4. Therefore, \(P(1)=P(2)=P(3)=P(4)=P(5)=0.125\) and \(P(6)=0.375\).

a) Use the equally likely outcomes model. \(P(A)=\frac{1}{2}=0.5\). b) \(P(1)=P(2)=P(3)=P(4)=P(5)=0.125\), and \(P(6)=0.375\).

- Count how many cards belong to each suit and each rank. - Equal numbers of cards produce equal probabilities. - Count face cards separately from all other cards.

1. \(S_1=\{\text{clubs},\text{diamonds},\text{hearts},\text{spades}\}\). Each suit has \(13\) cards, so each outcome has probability \(\frac{13}{52}=\frac{1}{4}\). 2. \(S_2=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\). Each rank has \(4\) cards, so each outcome has probability \(\frac{4}{52}=\frac{1}{13}\). 3. Each suit occurs the same number of times in the deck, and each rank occurs the same number of times, so the outcomes within each sample space are equally likely. 4. There are \(12\) face cards and \(40\) non-face cards. Thus, \(P(\text{face card})=\frac{12}{52}=\frac{3}{13}\), while \(P(\text{not a face card})=\frac{40}{52}=\frac{10}{13}\). The outcomes are not equally likely.

a) \(S_1=\{\text{clubs},\text{diamonds},\text{hearts},\text{spades}\}\); each probability is \(\frac{1}{4}\). b) \(S_2=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\); each probability is \(\frac{1}{13}\). c) Each suit and each rank appears equally often in the deck. d) No. \(P(\text{face card})=\frac{3}{13}\) and \(P(\text{not a face card})=\frac{10}{13}\).

- Interpret odds as relative numbers of event and nonevent cases. - The probability denominator includes both parts of the odds. - The complement exchanges the two roles.

1. Odds of \(3\) to \(5\) represent \(3\) event parts and \(5\) nonevent parts, for \(8\) total parts. 2. Thus \(P(A)=\frac38\) and \(P(A^c)=\frac58\). 3. The complement reverses the odds, so its odds are \(5\) to \(3\). 4. In general, odds of \(a\) to \(b\) give \(P(A)=\frac{a}{a+b}\).

a) \(P(A)=\frac38\), \(P(A^c)=\frac58\). b) The odds in favor of \(A^c\) are \(5\) to \(3\). c) Odds of \(a\) to \(b\) correspond to probability \(\frac{a}{a+b}\).

- Distinguish a probability from a guaranteed proportion in every finite block. - Ask whether outcomes other than three oversized parts are impossible or merely less likely. - Interpret the model across many repetitions rather than as a production quota.

1. The claim is false. A probability of \(0.30\) specifies the chance on each cycle and the long-run relative frequency under repeated independent cycles. 2. It does not require every short block to match the probability exactly. 3. A block with no oversized parts and a block with all ten oversized parts both have positive probability, although each is unlikely. Many other counts, including three, are possible.

The claim is false. The value \(0.30\) describes chance per cycle and long-run relative frequency, not an exact requirement of three oversized parts in every ten cycles. For example, a block could contain zero or ten oversized parts.

- Clarify what a single outcome looks like in this experiment. - Use complements when the event is easier to describe through what does not happen. - Use the structure of the sample space to justify the result.

1. \(A=\{3,6,9,12,15,18\}\) and \(B=\{13,14,15,16,17,18,19,20\}\). 2. \(A\cap B=\{15,18\}\). 3. \(A\cup B=\{3,6,9,12,13,14,15,16,17,18,19,20\}\). 4. \(A^c=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}\).

\(A\cap B=\{15,18\}\), \(A\cup B=\{3,6,9,12,13,14,15,16,17,18,19,20\}\), and \(A^c=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}\).

- The event depends only on the relative order of three named runners. - The unnamed runners may be inserted anywhere without favoring one of the named runners. - Use symmetry among the three possible runners who can lead the trio.

1. Ignore the positions of the other three runners and consider only the relative order of Maya, Leo, and Nia. 2. Each of the three runners is equally likely to be first among this trio. 3. Maya finishes ahead of both others exactly when she is first among the trio, so the probability is \(\frac13\).

The probability is \(\frac{1}{3}\).

- Think of an event as a selection from the listed outcomes. - For each outcome, what two inclusion choices are available? - To count events of a fixed size, focus on which outcomes are selected.

1. An event is any subset of the four-outcome sample space. 2. Each outcome is either included or not included, so there are \(2^4=16\) events. 3. Events with exactly two outcomes are the two-element subsets, so there are \(\binom42=6\). 4. An event is a collection of outcomes, not a new elementary result.

a) \(16\) b) \(6\) c) An event is a subset of \(S\), so it groups elementary outcomes rather than adding another outcome.

- Form the union before taking its complement. - Compute each complement relative to the full sample space. - Compare membership element by element.

1. \(A\cup B=\{a,c,d\}\), so \((A\cup B)^c=\{b,e\}\). 2. \(A^c=\{b,d,e\}\) and \(B^c=\{a,b,e\}\). 3. Their intersection is \(\{b,e\}\), equal to \((A\cup B)^c\).

\((A\cup B)^c=\{b,e\}\) and \(A^c\cap B^c=\{b,e\}\); the sets are equal.

- How do you find an interval probability for a continuous uniform distribution? - Which original values round to a given value with one decimal place? - What is the distance between consecutive numbers written to \(k\) decimal places? - How does that distance change when one more decimal place is used?

1. For a uniform distribution, the probability is the interval width divided by the total width: \(P(1.15\le X<1.25)=\frac{1.25-1.15}{5}=\frac{0.10}{5}=0.02\). 2. A value rounds to \(1.2\) to one decimal place when it lies in \([1.15,1.25)\). Therefore, \(P(\text{rounds to }1.2)=0.02\). 3. The values that round to \(a\) lie in \([a-0.5\cdot 10^{-k},a+0.5\cdot 10^{-k})\), an interval of width \(10^{-k}\). Thus, \(P=\frac{10^{-k}}{5}=0.2\cdot 10^{-k}\). Each additional decimal place divides the interval width, and therefore the probability, by \(10\).

1. \(P(1.15\le X<1.25)=0.02\) 2. \(P(\text{rounds to }1.2)=0.02\) 3. \(P=\frac{10^{-k}}{5}=0.2\cdot 10^{-k}\). The probability is proportional to the width \(10^{-k}\) of the rounding interval.

- What is the area over an interval whose width is exactly \(0\)? - Recall how probability is defined using a density function. - Compare the length of a line segment with the length of one point on that segment.

1. a) For any point \(c\), \(P(X=c)=\int_c^c f(x)\,dx=0\). Therefore, \(P(a\le X\le b)=P(X=a)+P(a<X<b)+P(X=b)=P(a<X<b)\). 2. b) For a continuous random variable, probability is not found by a countable sum of point probabilities. It is found by an integral, which is a limit of area sums and may be positive over an interval even though each individual point has probability \(0\). Similarly, a line segment has positive length even though each point has length \(0\).

a) Since \(P(X=c)=\int_c^c f(x)\,dx=0\), the endpoints \(a\) and \(b\) do not change the interval probability. b) Continuous probabilities are defined by integrals, not by adding individual point probabilities. An integral over an interval can be positive even though every point has probability \(0\).

- Think of the five drawings as five positions. - First choose the positions won by Alex, then choose the position won by Jordan. - What is the probability that a winner is neither Alex nor Jordan? - Find the probability of one specific qualifying order, then account for all such orders.

1. On each drawing, \(P(A)=\frac{1}{20}\), \(P(J)=\frac{1}{20}\), and the probability that someone else wins is \(P(O)=\frac{18}{20}\). 2. Choose the \(2\) drawings won by Alex and then the \(1\) drawing won by Jordan: \(\binom{5}{2}\binom{3}{1}=10\cdot 3=30\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{20}\right)^2\left(\frac{1}{20}\right)\left(\frac{18}{20}\right)^2\). 4. Therefore, the required probability is \(30\left(\frac{1}{20}\right)^3\left(\frac{18}{20}\right)^2=\frac{243}{80{,}000}=0.0030375\).

\(\frac{243}{80{,}000}=0.0030375\), or \(0.30375\%\)

- How many yellow draws must occur? - Find the probability of one specific order, such as red, red, blue, blue, yellow, yellow. - How many ways can the positions of the three colors be assigned? - Replacement makes the draws independent and keeps the color probabilities constant.

1. The color probabilities on each draw are \(P(R)=0.2\), \(P(B)=0.3\), and \(P(Y)=0.5\). 2. If exactly \(2\) draws are red and exactly \(2\) are blue, the remaining \(2\) draws must be yellow. 3. Choose the \(2\) red positions and then the \(2\) blue positions: \(\binom{6}{2}\binom{4}{2}=15\cdot 6=90\) color sequences. 4. Each qualifying sequence has probability \((0.2)^2(0.3)^2(0.5)^2=0.0009\). 5. Therefore, the probability is \(90\cdot 0.0009=0.081\).

\(0.081\), or \(8.1\%\)

- First find the probability of each number on one roll. - For a pair of different numbers, account for both possible orders. - Check that the probabilities add to \(1\).

1. The unordered outcomes are \(S=\{(1, 1),(1, 2),(1, 3),(2, 2),(2, 3),(3, 3)\}\). 2. On one roll, \(P(1)=\frac{1}{2}\), \(P(2)=\frac{1}{3}\), and \(P(3)=\frac{1}{6}\). 3. Therefore, \(P((1, 1))=\frac{1}{4}\), \(P((1, 2))=2\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{3}\), and \(P((1, 3))=2\left(\frac{1}{2}\right)\left(\frac{1}{6}\right)=\frac{1}{6}\). 4. Also, \(P((2, 2))=\frac{1}{9}\), \(P((2, 3))=2\left(\frac{1}{3}\right)\left(\frac{1}{6}\right)=\frac{1}{9}\), and \(P((3, 3))=\frac{1}{36}\). 5. These probabilities are not all equal, so assigning probability \(\frac{1}{6}\) to each unordered pair would be incorrect.

a) \(S=\{(1, 1),(1, 2),(1, 3),(2, 2),(2, 3),(3, 3)\}\) b) \(P((1, 1))=\frac{1}{4}\), \(P((1, 2))=\frac{1}{3}\), \(P((1, 3))=\frac{1}{6}\), \(P((2, 2))=\frac{1}{9}\), \(P((2, 3))=\frac{1}{9}\), and \(P((3, 3))=\frac{1}{36}\) c) The outcomes are not equally likely because their probabilities range from \(\frac{1}{36}\) to \(\frac{1}{3}\).

- Find the probability of each of the three result categories on one roll. - Choose the positions of the two \(6\)s first. - After those positions are fixed, how many positions remain for the three \(1\)s? - Multiply the number of qualifying arrangements by the probability of one arrangement.

1. On each roll, \(P(6)=\frac{1}{6}\), \(P(1)=\frac{1}{6}\), and \(P(\text{other})=\frac{4}{6}=\frac{2}{3}\). 2. Choose the \(2\) positions for the \(6\)s and then the \(3\) positions for the \(1\)s: \(\binom{10}{2}\binom{8}{3}=45\cdot 56\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^3\left(\frac{2}{3}\right)^5\). 4. Therefore, \(P=\binom{10}{2}\binom{8}{3}\left(\frac{1}{6}\right)^5\left(\frac{2}{3}\right)^5=\frac{280}{6561}\approx 0.0427\).

\(\frac{280}{6561}\approx 0.0427\), or about \(4.27\%\)

- Match each probability base with its exponent. - A binomial coefficient can choose positions for one category. - Why are only \(4\) positions available after the first choice?

1. The exponents show that event \(E\) contains exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package. 2. The factor \(\binom{12}{8}\) chooses the \(8\) positions occupied by light packages among the \(12\) selections. 3. After those positions are chosen, \(4\) positions remain. The factor \(\binom{4}{3}\) chooses the \(3\) positions occupied by standard packages. The final position must contain the heavy package.

Event \(E\) is “exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package.” The factor \(\binom{12}{8}\) chooses the light-package positions, and \(\binom{4}{3}\) chooses the standard-package positions from the remaining \(4\) positions.

- What values can \(|A|\) have compared with \(n\)? - How many outcomes are in the entire sample space? - If two sets have no elements in common, how many elements are in their union?

1. Since \(|A|\) is an integer from \(0\) to \(n\) and \(n>0\), \(P(A)=\frac{|A|}{n}\ge 0\). 2. Because \(|S|=n\), \(P(S)=\frac{|S|}{n}=\frac{n}{n}=1\). 3. If \(A\cap B=\emptyset\), then \(|A\cup B|=|A|+|B|\). Therefore, \(P(A\cup B)=\frac{|A|+|B|}{n}=\frac{|A|}{n}+\frac{|B|}{n}=P(A)+P(B)\).

1. \(|A|\ge 0\), so \(P(A)=\frac{|A|}{n}\ge 0\). 2. \(P(S)=\frac{n}{n}=1\). 3. For disjoint events, \(|A\cup B|=|A|+|B|\), so \(P(A\cup B)=P(A)+P(B)\).

- Recall the relationship between an event count and its relative frequency. - Can a count of observed outcomes be negative? - Across \(N\) trials, how many outcomes belong to the sample space? - If events are disjoint, can one trial be counted in both events?

1. The count \(H_N(A)\) is nonnegative. Since \(N>0\), \(h_N(A)=\frac{H_N(A)}{N}\ge 0\). 2. Every trial produces an outcome in \(S\), so \(H_N(S)=N\). Thus \(h_N(S)=\frac{N}{N}=1\). 3. If \(A\) and \(B\) are disjoint, no trial can be counted in both events. Therefore, \(H_N(A\cup B)=H_N(A)+H_N(B)\). Dividing by \(N\) gives \(h_N(A\cup B)=\frac{H_N(A)}{N}+\frac{H_N(B)}{N}=h_N(A)+h_N(B)\).

1. \(h_N(A)\ge 0\) because \(H_N(A)\ge 0\) and \(N>0\). 2. \(H_N(S)=N\), so \(h_N(S)=1\). 3. For disjoint events, \(H_N(A\cup B)=H_N(A)+H_N(B)\), so dividing by \(N\) proves additivity.

- How can you write the entire sample space using an event and its complement? - Which axiom gives the probability of the entire sample space? - When \(A\subseteq B\), split \(B\) into two disjoint parts. - Which axiom guarantees that a probability cannot be negative?

1. The events \(A\) and \(A^c\) are disjoint and satisfy \(A\cup A^c=S\). By additivity, \(P(S)=P(A)+P(A^c)\). 2. Since \(P(S)=1\), \(P(A^c)=1-P(A)\). 3. If \(A\subseteq B\), then \(B=A\cup(B\setminus A)\), and the two sets on the right are disjoint. 4. By additivity, \(P(B)=P(A)+P(B\setminus A)\). 5. Nonnegativity gives \(P(B\setminus A)\ge 0\), so \(P(B)\ge P(A)\).

a) \(P(A^c)=1-P(A)\) b) \(P(B)=P(A)+P(B\setminus A)\). Since \(P(B\setminus A)\ge 0\), \(P(A)\le P(B)\).

- What do the probability axioms require about the total probability and the sign of each probability? - Express each event probability as a sum of individual outcome probabilities. - Set up equations for the three unknown probabilities. - Check whether every resulting probability lies between \(0\) and \(1\).

1. Since \(\{e_1,e_2\}\) and \(\{e_3\}\) are disjoint and their union is \(S\), \(0.3+P(e_3)=1\). Thus \(P(e_3)=0.7\). 2. Since \(P(\{e_2,e_3\})=P(e_2)+P(e_3)\), \(0.6=P(e_2)+0.7\), so \(P(e_2)=-0.1\). 3. Since \(P(\{e_1,e_2\})=P(e_1)+P(e_2)\), \(0.3=P(e_1)-0.1\), so \(P(e_1)=0.4\). 4. The values add to \(1\), but \(P(e_2)=-0.1<0\). This violates nonnegativity, so no such probability distribution exists.

No such probability distribution exists. The proposed values imply \(P(e_1)=0.4\), \(P(e_2)=-0.1\), and \(P(e_3)=0.7\). The negative value violates the nonnegativity axiom.

- Separate the source of a probability value from the rules that the value must obey. - Which statement relies on symmetry, and which relies on observations? - Could a machine's defect rate be estimated without collecting production data?

1. Statement I uses an equally likely outcomes model because a fair four-sided die has four equally likely faces. The prime outcomes are \(2\) and \(3\), so the probability is \(\frac{2}{4}=0.5\). 2. Statement II is empirical because the value comes from long-term observed production data. 3. Statement III gives the mathematical rules that every valid probability model must satisfy, regardless of whether its numerical values come from symmetry, data, or another justified model.

a) I uses an equally likely outcomes model because the die is fair and \(2\) of the \(4\) outcomes are prime. II is empirical because the value is estimated from observed production data. b) The probability axioms provide the common mathematical framework that both approaches must satisfy; they do not specify how the numerical probabilities are obtained.

- What must the probabilities of all eight cards add to? - Write one equation that includes all four numbered cards and all four other cards. - Which cards are face cards? - In an equally likely outcomes model, compare favorable outcomes with all possible outcomes.

1. The probabilities must add to \(1\): \(4p+4(1.5p)=1\). Thus \(10p=1\), so \(p=0.10\). 2. Each face card has probability \(1.5\cdot 0.10=0.15\). There are three face cards, so \(P(E)=3\cdot 0.15=0.45\). 3. If all eight cards were equally likely, \(P(E)=\frac{3}{8}=0.375\). The weighted model gives a probability that is \(0.45-0.375=0.075\) greater.

a) \(p=0.10\) b) \(P(E)=0.45\) c) The equally likely outcomes model gives \(P(E)=\frac{3}{8}=0.375\), so the weighted model is higher by \(0.075\).

- List the perfect squares from \(1\) through \(10\). - Relative frequency is the number of occurrences divided by the number of trials. - The probabilities of all possible outcomes must add to \(1\). - Add the probabilities of the odd-numbered outcomes.

1. The perfect squares from \(1\) through \(10\) are \(1\), \(4\), and \(9\). Therefore, \(P(\text{perfect square})=\frac{3}{10}=0.3\). 2. The relative frequency is \(\frac{65}{500}=0.13\). For an equal-section spinner, \(P(10)=\frac{1}{10}=0.1\), so the observed relative frequency is \(0.03\) higher. 3. The probabilities must satisfy \(9q+0.2=1\). Thus, \(q=\frac{0.8}{9}=\frac{4}{45}\). There are \(5\) odd numbers among \(1\) through \(9\), so \(P(\text{odd})=5q=\frac{4}{9}\approx 0.444\).

1. \(P(\text{perfect square})=0.3\) 2. The relative frequency is \(0.13\), which is \(0.03\) higher than the theoretical probability \(0.1\). 3. \(q=\frac{4}{45}\), and \(P(\text{odd})=\frac{4}{9}\approx 0.444\).

- The probabilities of all outcomes in a sample space must add to \(1\). - Compare an equally likely-outcomes model with the given unequal probability model. - Compare the observed relative frequency of \(6\) with its theoretical probability in each model.

1. The probabilities must add to \(1\), so \(3\cdot0.1+3x=1\). Therefore, \(3x=0.7\) and \(x=\frac{7}{30}\approx 0.233\). 2. Since \(E=\{4,5,6\}\), the first model gives \(P(E)=\frac{7}{30}+0.1+\frac{7}{30}=\frac{17}{30}\approx 0.567\). Under the equally likely-outcomes model, \(P(E)=\frac{3}{6}=0.5\). 3. The relative frequency is \(\frac{140}{600}=\frac{7}{30}\approx 0.233\). This equals \(P(6)\) in the first model, while the equally likely-outcomes model gives \(P(6)=\frac{1}{6}\approx 0.167\). Therefore, the first model better matches this observation. One outcome frequency alone does not validate the entire model.

1. \(x=\frac{7}{30}\approx 0.233\) 2. First model: \(P(E)=\frac{17}{30}\approx 0.567\); equally likely-outcomes model: \(P(E)=0.5\) 3. The first model better matches the observed frequency of \(6\), but this observation alone does not confirm the complete model.

- Use ordered pairs for the two rolls. - Count how many ordered pairs produce each sum. - Check nonnegativity and that the probabilities total \(1\).

1. \(S_A=\{(i, j): i, j\in\{1, 2, 3, 4\}\}\), so \(|S_A|=4^2=16\). The die is fair and the rolls are independent, so each ordered pair has probability \(\frac{1}{16}\). 2. \(S_B=\{2,3,4,5,6,7,8\}\). Counting ordered pairs gives probabilities \(\frac{1}{16},\frac{2}{16},\frac{3}{16},\frac{4}{16},\frac{3}{16},\frac{2}{16},\frac{1}{16}\), respectively. 3. Every probability is nonnegative, and their sum is \(\frac{1+2+3+4+3+2+1}{16}=1\). Event probabilities formed by adding the relevant outcome probabilities are additive for disjoint events. 4. In \(S_A\), every outcome is equally likely, so probabilities can be found by counting favorable ordered pairs. The unequal probabilities of sums then follow from how many ordered pairs produce each sum.

a) \(S_A=\{(i, j): i, j\in\{1, 2, 3, 4\}\}\), \(|S_A|=16\), and every pair has probability \(\frac{1}{16}\). b) For sums \(2,3,4,5,6,7,8\), the probabilities are \(\frac{1}{16},\frac{2}{16},\frac{3}{16},\frac{4}{16},\frac{3}{16},\frac{2}{16},\frac{1}{16}\). c) Yes. The probabilities are nonnegative and sum to \(1\). d) The equally likely ordered pairs make probability calculations a counting problem.

- Multiply the probability of reaching a folder by the conditional probability of a file within it. - Equal choices at each stage do not imply equal probabilities for all terminal outcomes. - Check whether the two events in part c can occur together.

1. Each A-file has probability \(\frac12\cdot\frac12=\frac14\). Each B-file has probability \(\frac12\cdot\frac14=\frac18\). 2. Uniformity occurs within each stage, but the second-stage denominators differ by folder. A named file in the smaller folder receives more probability. 3. “Folder B” and “file \(a_1\)” are disjoint. Their probabilities are \(\frac12\) and \(\frac14\), so the union has probability \(\frac34\).

a) \(P(a_1)=P(a_2)=\frac14\), and \(P(b_1)=P(b_2)=P(b_3)=P(b_4)=\frac18\). b) Files inherit different path probabilities because the folders contain different numbers of files. c) The probability is \(\frac34\).

- Evaluate each forecast level only among the days receiving that forecast. - Calibration compares forecast probabilities with observed relative frequencies. - Compare error magnitudes rather than only their signs.

1. For the \(70\%\) group, the empirical frequency is \(\frac{136}{200}=0.68\), giving calibration error \(0.68-0.70=-0.02\). 2. For the \(40\%\) group, the empirical frequency is \(\frac{35}{80}=0.4375\), giving error \(0.4375-0.40=0.0375\). 3. Absolute errors are \(0.02\) and \(0.0375\), so the \(70\%\) forecasts are better calibrated in this data set.

a) Empirical frequencies: \(0.68\) and \(0.4375\). b) Calibration errors: \(-0.02\) and \(0.0375\). c) The \(70\%\) forecast group has the smaller absolute calibration error.

- In a uniform model, event probabilities are integer counts divided by the sample-space size. - The sample-space size must accommodate both denominators. - After finding the minimum size, assign nonoverlapping blocks of outcomes.

1. In a uniform sample space of size \(N\), the event sizes must be \(\frac{N}{3}\) and \(\frac{N}{4}\), so \(N\) must be divisible by both \(3\) and \(4\). 2. The smallest such size is \(\operatorname{lcm}(3,4)=12\). 3. Let \(S=\{1,2,\ldots,12\}\), \(A=\{1,2,3,4\}\), and \(B=\{5,6,7\}\). Then the events are disjoint and have probabilities \(\frac{4}{12}=\frac13\) and \(\frac{3}{12}=\frac14\). 4. No smaller \(N\) is divisible by both denominators, so no smaller uniform construction is possible.

The minimum sample-space size is \(12\). One construction is \(S=\{1,2,\ldots,12\}\), \(A=\{1,2,3,4\}\), and \(B=\{5,6,7\}\). No smaller uniform sample space works because its size would have to be divisible by both \(3\) and \(4\).

- Compare the two interval endpoints \(\frac{1}{n+1}\) and \(\frac1n\). - Uniform probability on \((0,1)\) is determined by interval length. - Ask whether any fixed positive value can remain below every reciprocal \(\frac1n\).

1. The event is \(A_n=\{0<X<\frac1n\}\). Since \(\frac{1}{n+1}<\frac1n\), every outcome in \(A_{n+1}\) is also in \(A_n\). 2. Uniform probability on \((0,1)\) gives \(P(A_n)=\frac1n\). 3. No positive number is smaller than \(\frac1n\) for every positive integer \(n\), so the event common to all \(A_n\) is \(\{\}\). 4. Therefore the probability of the common event is \(0\), and \(P(A_n)=\frac1n\) approaches \(0\) as \(n\) increases without bound.

a) Every outcome in \(A_{n+1}\) is also in \(A_n\). b) \(P(A_n)=\frac{1}{n}\). c) The event common to every \(A_n\) is \(\{\}\). Its probability is \(0\), which equals the value approached by \(P(A_n)\).

- Which individual standard places the strongest limit on the maximum? - Consider the probabilities of failing each standard. - What happens to the minimum number of perfect parts if the three types of failures occur on different parts?

1. The probability of meeting all three standards cannot exceed the smallest individual probability, so \(P(A\cap B\cap C)\le 0.80\). 2. Let \(A^c\), \(B^c\), and \(C^c\) represent failure to meet each standard. Their probabilities are \(0.10\), \(0.15\), and \(0.20\). 3. The probability of failing at least one standard is at most \(0.10+0.15+0.20=0.45\). Therefore, the probability of meeting all three is at least \(1-0.45=0.55\). 4. Both bounds are attainable, so \(P(A\cap B\cap C)\in[0.55,0.80]\).

The probability is at least \(55\%\) and at most \(80\%\).

- A bijection preserves the number of outcomes in a finite set. - Uniform probability depends only on event cardinality. - For the application, relabel people while leaving the ordering structure unchanged.

1. A bijection maps distinct elements of \(A\) to distinct elements of \(B\) and reaches every element of \(B\), so \(|A|=|B|\). 2. In a uniform finite model, \(P(A)=\frac{|A|}{|S|}=\frac{|B|}{|S|}=P(B)\). 3. Relabeling any one person as any other gives a bijection between orderings in which the first person wins and orderings in which the second person wins. Thus all four first-place events have equal probability. 4. These four events partition the sample space, so each probability is \(\frac14\).

a) \(P(A)=P(B)\). b) A probability-preserving relabeling pairs the outcomes of symmetric events one-to-one. c) Each person finishes first with probability \(\frac14\).

- A convex combination uses nonnegative weights that sum to one. - Check each probability axiom by distributing the weighted sum. - Compute atom probabilities before combining them into the requested event.

1. Nonnegativity is preserved because \(R(E)\) is a nonnegative weighted average. Also \(R(S)=0.40\cdot1+0.60\cdot1=1\). 2. For disjoint events, additivity of \(P\) and \(Q\) passes through the weighted sum, so \(R\) is a probability model. 3. Atom probabilities are \(R(a)=0.40\cdot0.50+0.60\cdot0.10=0.26\), \(R(b)=0.40\cdot0.30+0.60\cdot0.20=0.24\), and \(R(c)=0.40\cdot0.20+0.60\cdot0.70=0.50\). 4. \(R(\{a,c\})=0.26+0.50=0.76\).

a) \(R\) satisfies the probability axioms. b) \(R=(0.26,0.24,0.50)\). c) \(R(\{a,c\})=0.76\).