Aimathic – Geometric distribution

Geometric distribution

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Difficulty: 1 – 5

53089112

A data-center security system has three detection stages in sequence. Each stage independently detects an unauthorized access attempt with probability $0.80$. The attempt is stopped as soon as one stage detects it. Find the probability that an unauthorized attempt is 1. stopped by the first stage; 2. first stopped by the third stage; 3. detected by none of the three stages.

Hints

- What is the complement of a stage detecting the attempt? - What must happen at the first two stages for the third stage to be the first detector? - Which rule applies when independent events must occur in sequence?

Solution

1. Let $p=0.80$ be the detection probability and $q=1-p=0.20$ be the miss probability. 2. The first stage stops the attempt with probability $p=0.80$. 3. For the third stage to be the first to detect the attempt, the first two stages must miss and the third must detect it. The probability is $q^2p=(0.20)^2(0.80)=0.032$. 4. If no stage detects the attempt, all three stages miss. The probability is $q^3=(0.20)^3=0.008$.

Answer

1. $0.80$ 2. $0.032$ 3. $0.008$

53752612

Each quality inspection detects a defect with probability $0.3$. The inspections are independent. Find the probability that the first detected defect occurs on the third inspection.

Hints

- Identify the exact sequence required for the first detection to occur on inspection $3$. - Multiply the probabilities along that one sequence.

Solution

1. The required sequence is no defect, no defect, defect. 2. Its probability is $P(X=3)=(0.7)^2\cdot 0.3=0.147$.

Answer

$0.147=14.7\%$

54746812

The number of auditions through the first callback is modeled by a geometric random variable $X$ with mean $5$. A student says the standard deviation must also be $5$ because geometric distributions are strongly right-skewed. Determine the success probability, find the actual standard deviation, and evaluate the student's claim.

Hints

- Use the given long-run average wait to recover the trial success rate. - Treat center and spread as different properties of the distribution. - Check the claim numerically rather than relying on the shape alone.

Solution

1. Since $E(X)=\frac{1}{p}=5$, the success probability is $p=0.20$. 2. The standard deviation is $\sigma_X=\frac{\sqrt{1-p}}{p}=\frac{\sqrt{0.80}}{0.20}=\sqrt{20}\approx4.47$. 3. The student's claim is false; skewness does not force the mean and standard deviation to be equal.

Answer

$p=0.20$ and $\sigma_X=\sqrt{20}\approx4.47$. The student's claim is false.

54747112

A device is tested repeatedly. On odd-numbered tests, the probability of passing is $0.20$; on even-numbered tests, the probability of passing is $0.50$. Test outcomes are independent. Let $X$ be the test number of the first pass. Is $X$ geometrically distributed? Explain, and find $P(X=4)$.

Hints

- Compare the trial structure with all requirements of a geometric model. - List the outcome needed on each of the first four tests. - Use the stated independence after assigning the correct probability to each position.

Solution

1. A geometric model requires the same success probability on every trial. Here the probability alternates, so $X$ is not geometric. 2. For $X=4$, tests $1$, $2$, and $3$ must fail and test $4$ must pass. 3. The probability is $(0.80)(0.50)(0.80)(0.50)=0.16$.

Answer

No. The success probability is not constant. $P(X=4)=0.16$.

54747512

A geometric process has success probability $0.27$ on each trial. To simulate it with two-digit random integers from $00$ through $99$, a student labels $00$ through $27$ as success and all other values as failure. a) Explain the student's error and give a correct assignment. b) Using the intended model, find the probability that more than $3$ trials are needed.

Hints

- Count the endpoints in the proposed integer interval carefully. - Match the number of success labels to the intended percentage. - Interpret a wait longer than three trials as an initial run of failures.

Solution

1. The integers $00$ through $27$ include $28$ outcomes, so the student's assignment gives success probability $0.28$. 2. One correct assignment is $00$ through $26$ for success and $27$ through $99$ for failure. 3. More than three trials means three consecutive failures, so the probability is $(0.73)^3\approx0.3890$.

Answer

a) The student assigned $28$ of the $100$ integers to success. A correct assignment is $00$–$26$ for success and $27$–$99$ for failure. b) $P(X>3)=(0.73)^3\approx0.3890$.

54747812

A quality check passes independently with probability $0.45$ on each attempt. Let $F$ be the number of failed checks before the first pass. Find the mean and standard deviation of $F$.

Hints

- Identify whether the successful trial is included in the variable. - A one-unit shift changes the center but not the spread. - Keep the failure probability separate from the success probability.

Solution

1. The variable counts failures before success, so its mean is $E(F)=\frac{1-p}{p}=\frac{0.55}{0.45}\approx1.22$. 2. Shifting a trials-through-success variable down by one does not change standard deviation. 3. Therefore, $\sigma_F=\frac{\sqrt{1-p}}{p}=\frac{\sqrt{0.55}}{0.45}\approx1.65$.

Answer

$E(F)\approx1.22$ failures and $\sigma_F\approx1.65$ failures.

54748812

A geometric random variable $X$ counts attempts through the first success, with $p=0.28$. A student writes $P(X\ge5)=(0.72)^5$. Explain the error and find the correct probability.

Hints

- List the smallest waiting time included in the event. - Determine exactly which earlier trials must have failed. - Keep the difference between “at least” and “more than” clear.

Solution

1. The event $X\ge5$ means that no success occurs on attempts $1$ through $4$. 2. It requires four failures, not five failures. 3. Therefore, $P(X\ge5)=(0.72)^4\approx0.2687$.

Answer

The student counted one extra failure. The correct probability is $P(X\ge5)=(0.72)^4\approx0.2687$.

54749212

A geometric random variable $X$ counts trials through the first success, with $p=0.05$. Its mean is $20$ trials. Find the probability that the actual wait is longer than its mean, and interpret the result.

Hints

- Translate “longer than the mean” into a tail event using the integer-valued variable. - Determine how many initial failures that event requires. - Interpret the probability in light of the distribution's shape.

Solution

1. A wait longer than the mean is the event $X>20$. 2. This requires the first $20$ trials to fail, so $P(X>20)=(0.95)^{20}\approx0.3585$. 3. Thus, even though the mean wait is $20$, about $35.85\%$ of waits exceed $20$ trials because the distribution has a long right tail.

Answer

$P(X>20)=(0.95)^{20}\approx0.3585$. About $35.85\%$ of waits are longer than the mean.

54749512

A geometric random variable $X$ has mean $8$. a) Find the success probability. b) Compare $P(X=1)$ with $P(X=8)$. c) Explain why the most likely value can be much smaller than the mean.

Hints

- Recover the trial success rate from the given center. - Evaluate each requested point probability separately. - Distinguish the most likely single value from the balance point of the entire distribution.

Solution

1. Since $E(X)=\frac{1}{p}=8$, $p=0.125$. 2. $P(X=1)=0.125$. 3. $P(X=8)=(0.875)^7(0.125)\approx0.0491$. 4. Geometric point probabilities decrease as the trial number increases, so the mode is $1$. Rare long waits pull the mean to the right.

Answer

a) $p=0.125$. b) $P(X=1)=0.125$, while $P(X=8)\approx0.0491$. c) The distribution is right-skewed; long waits raise the mean even though trial $1$ is the single most likely value.

52363512

A spinner has five equal sections, and exactly one section is blue. The spinner is spun until blue appears for the first time. Let $X$ be the number of spins required. a) Find $P(X=2)$ and $P(X=3)$. b) Find the probability that at most three spins are required. c) Find the probability that more than four spins are required.

Hints

- For $X=k$, identify the failures before the first success. - “At most three” includes success on spin $1$, $2$, or $3$. - For part c, determine what must happen on the first four spins.

Solution

1. The success probability is $p=\frac{1}{5}=0.20$, and the failure probability is $0.80$. 2. $P(X=2)=0.80\cdot 0.20=0.16$. 3. $P(X=3)=(0.80)^2\cdot 0.20=0.128$. 4. $P(X\le 3)=1-P(\text{no blue in the first three spins})=1-(0.80)^3=0.488$. 5. More than four spins are required exactly when the first four spins are not blue, so $P(X>4)=(0.80)^4=0.4096$.

Answer

a) $P(X=2)=0.16$; $P(X=3)=0.128$ b) $P(X\le 3)=0.488$ c) $P(X>4)=0.4096$

52363612

At a factory, each component independently has probability $0.25$ of needing rework. Let $X$ be the number of components inspected until the first component needing rework is found. Find the smallest number $n$ of inspections for which the probability of finding at least one component needing rework is at least $80\%$.

Hints

- Use the complement of finding at least one component needing rework. - Write an inequality involving the probability of no successes in $n$ trials. - Test consecutive whole-number values near the threshold.

Solution

1. The probability that a component does not need rework is $0.75$. 2. The probability of finding at least one component needing rework in $n$ inspections is $P(X\le n)=1-(0.75)^n$. 3. Require $1-(0.75)^n\ge 0.80$, or $(0.75)^n\le 0.20$. 4. For $n=5$, $1-(0.75)^5\approx 0.7627$, which is less than $0.80$. 5. For $n=6$, $1-(0.75)^6\approx 0.8220$, which meets the requirement. Therefore, the smallest value is $n=6$.

Answer

$n=6$

53098512

An online retailer sends a promotional email to a new customer. The probability that the customer opens any one email is $12\%$, independently of the other emails. The retailer sends one email per day until the customer opens one, but sends at most $5$ emails total. Find the expected number of emails the customer receives.

Hints

- List the possible numbers of emails the customer can receive. - A fifth email is sent whenever the first four are not opened. - For fewer than five emails, use the probability that the first opening occurs on that email. - Multiply each possible value by its probability and add.

Solution

1. Let $X$ be the number of emails sent. Then $X\in\{1,2,3,4,5\}$, and the probability an email is not opened is $q=0.88$. 2. The probabilities are $P(X=1)=0.12$, $P(X=2)=0.88(0.12)=0.1056$, $P(X=3)=0.88^2(0.12)=0.092928$, $P(X=4)=0.88^3(0.12)=0.08177664$, and $P(X=5)=0.88^4=0.59969536$. The last probability includes every case in which the first four emails are not opened. 3. Therefore, $E(X)=1(0.12)+2(0.1056)+3(0.092928)+4(0.08177664)+5(0.59969536)=3.93556736$.

Answer

The customer receives an average of about $3.94$ emails.

53098612

At a carnival game, each independent spin has a $25\%$ chance of winning. A player spins until winning once, but stops after at most $4$ spins. Each spin costs $\$1.50$. Find the expected total cost of one series of spins.

Hints

- List the possible numbers of spins in one series. - The fourth spin occurs whenever the first three spins are losses. - Find the expected number of spins first. - Multiply the expected number of spins by the cost per spin.

Solution

1. Let $X$ be the number of spins. Then $X\in\{1,2,3,4\}$, with success probability $p=0.25$ and failure probability $q=0.75$. 2. The probabilities are $P(X=1)=0.25$, $P(X=2)=0.75(0.25)=0.1875$, $P(X=3)=0.75^2(0.25)=0.140625$, and $P(X=4)=0.75^3=0.421875$. 3. Thus, $E(X)=1(0.25)+2(0.1875)+3(0.140625)+4(0.421875)=2.734375$ spins. 4. The expected cost is $\$1.50\cdot2.734375=\$4.1015625$, which rounds to $\$4.10$.

Answer

The expected total cost is about $\$4.10$.

54725212

A fair coin is flipped until the first head appears. Let $T$ be the number of flips. a) Describe the infinite sample space using the possible values of $T$. b) Find $P(T=n)$ for any positive integer $n$. c) Verify that the probabilities of all possible stopping times sum to $1$.

Hints

- Describe what sequence of flips corresponds to stopping on flip $n$. - Multiply the probabilities of the required tail sequence and final head. - Recognize the total probability as an infinite geometric series.

Solution

1. The possible stopping times are $S=\{1,2,3,\ldots\}$. The outcome $T=n$ means $n-1$ tails followed by one head. 2. Independence gives $P(T=n)=\left(\frac12\right)^{n-1}\cdot\frac12=2^{-n}$. 3. The total is the geometric series $\sum_{n=1}^{\infty}2^{-n}=\frac{\frac12}{1-\frac12}=1$.

Answer

a) $S=\{1,2,3,\ldots\}$. b) $P(T=n)=2^{-n}$. c) $\sum_{n=1}^{\infty}2^{-n}=1$, so the model assigns total probability $1$.

54738112

An emergency beacon transmits successfully on each attempt with probability $0.65$. Attempts are independent. Find the smallest number of attempts needed so that the probability of at least one successful transmission is at least $0.99$.

Hints

- Work with the complementary event that every attempt fails. - Express the target as an inequality involving the number of attempts. - After solving the inequality, check the neighboring integers.

Solution

1. The probability that all $n$ attempts fail is $(1-0.65)^n=0.35^n$. 2. The requirement is $1-0.35^n\ge0.99$, so $0.35^n\le0.01$. 3. Taking logarithms gives $n\ge\frac{\ln(0.01)}{\ln(0.35)}\approx4.387$. 4. The smallest integer is $n=5$. Indeed, four attempts give $1-0.35^4=0.98499375<0.99$, while five give $1-0.35^5=0.9947478125\ge0.99$.

Answer

The beacon needs at least $5$ attempts.

54746212

A box contains $3$ green cards and $5$ white cards. Cards are drawn one at a time without replacement until the first green card appears. Let $X$ be the draw number of the first green card. A student claims that $X$ has a geometric distribution with success probability $\frac{3}{8}$. Is the student correct? Explain, and find $P(X=3)$.

Hints

- Check what happens to the chance of success after an unsuccessful draw. - Compare the sampling process with the conditions required for repeated identical trials. - Describe the exact sequence of outcomes needed for the first success on draw $3$.

Solution

1. The draws are not independent because cards are not replaced, and the probability of green changes after each white card. Therefore, $X$ is not geometric. 2. For $X=3$, the first two cards must be white and the third green. 3. The probability is $\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{5}{28}\approx0.1786$.

Answer

No. The trials are not independent and the success probability is not constant. $P(X=3)=\frac{5}{28}\approx0.1786$.

54746312

A network request succeeds independently on each attempt with probability $0.18$. Let $Y$ be the number of attempts through the first success, and let $X$ be the number of failed attempts before the first success. a) State the relationship between $X$ and $Y$. b) Find $P(X=4)$. c) Find the mean of each variable.

Hints

- Decide whether the successful trial is included in each count. - Describe the exact sequence represented by four failures before success. - Use the one-unit shift between the variables when comparing their centers.

Solution

1. Since the successful attempt is included in $Y$ but not in $X$, $Y=X+1$. 2. The event $X=4$ requires four failures followed by a success, so $P(X=4)=(0.82)^4(0.18)\approx0.0814$. 3. The mean number of attempts through success is $E(Y)=\frac{1}{0.18}\approx5.56$. 4. Therefore, $E(X)=E(Y)-1=\frac{0.82}{0.18}\approx4.56$.

Answer

a) $Y=X+1$. b) $P(X=4)=(0.82)^4(0.18)\approx0.0814$. c) $E(X)\approx4.56$ failures and $E(Y)\approx5.56$ attempts.

54746412

For a geometric random variable $X$ that counts trials through the first success, $P(X>5)=0.20$. Find the success probability $p$ for each trial and the mean of $X$.

Hints

- Translate the tail event into a sequence with no successes. - Use the repeated-trial structure to form an equation for the failure probability. - Once the trial success rate is known, interpret the long-run average count.

Solution

1. The event $X>5$ means the first five trials are failures, so $(1-p)^5=0.20$. 2. Thus, $1-p=0.20^{1/5}$, giving $p=1-0.20^{1/5}\approx0.2752$. 3. The mean is $E(X)=\frac{1}{p}\approx3.63$ trials.

Answer

$p=1-0.20^{1/5}\approx0.2752$, and $E(X)\approx3.63$ trials.

54746512

A rare signal is detected independently on each scan with probability $0.12$. Eight scans have already failed to detect it. a) Given this information, find the probability that more than $5$ additional scans will be needed. b) Given this information, find the expected number of additional scans through detection.

Hints

- Decide whether earlier failures change the chance on the next trial. - Reframe the question as a fresh waiting-time problem beginning now. - Distinguish a tail probability from a long-run average wait.

Solution

1. The geometric model is memoryless, so the eight previous failures do not change the distribution of the remaining wait. 2. More than five additional scans means five more failures, with probability $(0.88)^5\approx0.5277$. 3. The expected number of additional scans through detection is $\frac{1}{0.12}\approx8.33$.

Answer

a) $P(\text{more than 5 additional scans})=(0.88)^5\approx0.5277$. b) The expected additional wait is approximately $8.33$ scans.

54746712

A geometric random variable $X$ counts trials through the first success. It is known that $\frac{P(X=5)}{P(X=4)}=0.65$. Find the success probability $p$ and then find $P(X=4)$.

Hints

- Write expressions for the two adjacent point probabilities. - Simplify their ratio before substituting numbers. - Use the recovered trial rate in the probability for the requested count.

Solution

1. For a geometric variable, $P(X=5)=(1-p)^4p$ and $P(X=4)=(1-p)^3p$. 2. Their ratio is $1-p$, so $1-p=0.65$ and $p=0.35$. 3. Then $P(X=4)=(0.65)^3(0.35)\approx0.0961$.

Answer

$p=0.35$, and $P(X=4)\approx0.0961$.

54746912

A geometric random variable $X$ counts attempts through the first success, with success probability $0.40$. Given that the first success occurs within the first $4$ attempts, find the probability that it occurs on attempt $2$.

Hints

- Express the condition using a cumulative event. - Check whether the event in the numerator is contained in the conditioning event. - Compute the point probability and the cumulative probability separately.

Solution

1. The requested conditional probability is $P(X=2\mid X\le4)=\frac{P(X=2)}{P(X\le4)}$. 2. The numerator is $P(X=2)=(0.60)(0.40)=0.24$. 3. The denominator is $P(X\le4)=1-(0.60)^4=0.8704$. 4. Therefore, $P(X=2\mid X\le4)=\frac{0.24}{0.8704}\approx0.2757$.

Answer

$P(X=2\mid X\le4)\approx0.2757$.

54747012

A diagnostic test gives a usable reading independently on each attempt with probability $0.22$. Let $X$ be the attempt number of the first usable reading. Find $P(3\le X\le6)$.

Hints

- Translate the lower bound into a required string of early failures. - Remove outcomes in which the wait continues beyond the upper bound. - A difference of two tail probabilities can avoid adding several separate cases.

Solution

1. The event $3\le X\le6$ means the first two attempts fail, but not all of the first six attempts fail. 2. The probability of at least two initial failures is $(0.78)^2$. 3. Subtract the probability of six initial failures: $P(3\le X\le6)=(0.78)^2-(0.78)^6\approx0.3832$.

Answer

$P(3\le X\le6)=(0.78)^2-(0.78)^6\approx0.3832$.

54747312

A geometric random variable $X$ counts calls through the first answered call, with answer probability $0.25$ on each independent call. Given that the first $12$ calls were unanswered, find $P(X\le15\mid X>12)$.

Hints

- Reinterpret the condition as a new starting point after a string of failures. - Count how many additional trials are allowed. - Use the complement of no success in those additional trials.

Solution

1. Given $X>12$, the process restarts with the same success probability because of memorylessness. 2. The event $X\le15$ then means at least one answer in the next three calls. 3. The probability is $1-(0.75)^3=0.578125\approx0.5781$.

Answer

$P(X\le15\mid X>12)=1-(0.75)^3\approx0.5781$.

54747412

A contestant repeats an independent challenge until the first success. Each attempt costs $\$3.00$, the success probability is $0.16$ per attempt, and a successful completion pays a single prize of $\$40.00$. a) Find the expected total attempt cost. b) Find the expected net payoff. c) What prize would make the expected net payoff equal to $\$0$?

Hints

- First find the long-run average number of attempts through success. - Convert the expected attempt count into an expected total cost. - For break-even, compare the fixed prize with the expected cost.

Solution

1. The expected number of attempts is $\frac{1}{0.16}=6.25$. 2. The expected total attempt cost is $\$3.00(6.25)=\$18.75$. 3. The expected net payoff is $\$40.00-\$18.75=\$21.25$. 4. A break-even prize must equal the expected cost, so it is $\$18.75$.

Answer

a) $\$18.75$. b) $\$21.25$. c) $\$18.75$.

54747612

A rescue beacon has the same independent probability $p$ of connecting on each attempt. Let $X$ be the number of attempts through the first connection. Find the minimum value of $p$ that makes $P(X\le8)\ge0.95$.

Hints

- Rephrase connection within eight attempts using its complement. - Translate the probability requirement into an inequality for repeated failures. - Keep the direction of the inequality in mind when solving for the success rate.

Solution

1. The probability of at least one connection in eight attempts is $1-(1-p)^8$. 2. Require $1-(1-p)^8\ge0.95$, so $(1-p)^8\le0.05$. 3. Thus, $1-p\le0.05^{1/8}$, giving $p\ge1-0.05^{1/8}\approx0.3123$.

Answer

The minimum success probability is $p=1-0.05^{1/8}\approx0.3123$.

54747712

Two independent troubleshooting procedures are repeated until the first success. Procedure A succeeds with probability $0.10$ per attempt, and Procedure B succeeds with probability $0.20$ per attempt. Let $X_A$ and $X_B$ be the corresponding attempt counts. a) For each procedure, find the probability that the first success occurs on attempt $5$. b) For each procedure, find the probability that more than $5$ attempts are needed. c) Explain why the procedure with the larger success probability can have a larger probability of success on exactly attempt $5$ but a smaller probability of waiting beyond attempt $5$.

Hints

- Distinguish an exact attempt number from a wait that continues past that number. - Write the required success-failure sequence for the point probabilities. - Think about how increasing the trial success rate redistributes probability across all possible waits.

Solution

1. $P(X_A=5)=(0.90)^4(0.10)\approx0.0656$. 2. $P(X_B=5)=(0.80)^4(0.20)\approx0.0819$. 3. $P(X_A>5)=(0.90)^5\approx0.5905$. 4. $P(X_B>5)=(0.80)^5\approx0.3277$. 5. Procedure B moves probability toward earlier attempts. That can increase the mass at attempt $5$ while greatly reducing the entire tail beyond $5$.

Answer

a) Procedure A: approximately $0.0656$; Procedure B: approximately $0.0819$. b) Procedure A: approximately $0.5905$; Procedure B: approximately $0.3277$. c) The larger success rate shifts probability toward earlier attempts, affecting a single point and the later tail differently.

54748012

A search algorithm originally succeeds independently with probability $0.08$ per run. An update doubles the success probability to $0.16$. Let $X$ and $Y$ be the numbers of runs through the first success before and after the update. Compare the means and standard deviations of $X$ and $Y$. Does doubling $p$ exactly halve both measures?

Hints

- Evaluate center and spread separately at each success rate. - Compare the algebraic dependence of each measure on the trial probability. - Do not assume that the same proportional change applies to both measures.

Solution

1. Before the update, $E(X)=\frac{1}{0.08}=12.5$ and $\sigma_X=\frac{\sqrt{0.92}}{0.08}\approx11.99$. 2. After the update, $E(Y)=\frac{1}{0.16}=6.25$ and $\sigma_Y=\frac{\sqrt{0.84}}{0.16}\approx5.73$. 3. Doubling $p$ exactly halves the mean because the mean is $\frac{1}{p}$. It does not exactly halve the standard deviation because the numerator $\sqrt{1-p}$ also changes.

Answer

Before: mean $12.5$, standard deviation $\approx11.99$. After: mean $6.25$, standard deviation $\approx5.73$. The mean is exactly halved, but the standard deviation is not.

54748112

A geometric random variable $X$ counts trials through the first success, with $p=0.20$. a) Find the median of $X$, defined as the smallest integer $m$ for which $P(X\le m)\ge0.50$. b) Find $P(X=m)$.

Hints

- Translate the median definition into a cumulative probability threshold. - Because the variable is discrete, check the first integer that crosses the threshold. - After finding the median value, treat the second part as an exact waiting-time event.

Solution

1. Solve $1-(0.80)^m\ge0.50$, or $(0.80)^m\le0.50$. 2. This gives $m\ge\frac{\log(0.50)}{\log(0.80)}\approx3.11$, so the smallest integer is $m=4$. 3. The cumulative probability at $4$ is $1-(0.80)^4=0.5904$, while the cumulative probability at $3$ is below $0.50$. 4. $P(X=4)=(0.80)^3(0.20)=0.1024$.

Answer

a) The median is $m=4$. b) $P(X=m)=0.1024$.

54748412

A geometric process averages $3.5$ failures before the first success. Let $X$ count trials through the first success. a) Find the success probability $p$. b) Find $P(X\le4)$.

Hints

- Match the given average to the counting convention in the statement. - Solve for the trial success rate before addressing the cumulative event. - Use the complement of four initial failures.

Solution

1. The expected number of failures before success is $\frac{1-p}{p}=3.5$. 2. Solving $1-p=3.5p$ gives $p=\frac{2}{9}$. 3. The failure probability is $\frac{7}{9}$, so $P(X\le4)=1-\left(\frac{7}{9}\right)^4\approx0.6340$.

Answer

a) $p=\frac{2}{9}\approx0.2222$. b) $P(X\le4)\approx0.6340$.

54748612

Two independent detectors scan once per round. Detector A finds a signal with probability $0.20$ per round, and detector B finds it with probability $0.30$ per round. Let $M$ be the first round in which at least one detector finds the signal. a) Explain why $M$ is geometric and find its success probability. b) Find $E(M)$. c) Find $P(M=3)$.

Hints

- Define success for one combined round. - Use the chance that both detectors miss to find the combined success rate. - Once the round-level model is established, treat the rounds as repeated trials.

Solution

1. A round fails only if both detectors fail, with probability $(0.80)(0.70)=0.56$. 2. Therefore, a round succeeds with probability $1-0.56=0.44$, and identical independent rounds make $M$ geometric. 3. The mean is $E(M)=\frac{1}{0.44}\approx2.2727$ rounds. 4. The probability of first success on round $3$ is $(0.56)^2(0.44)\approx0.1380$.

Answer

a) $M$ is geometric with $p=0.44$. b) $E(M)\approx2.2727$ rounds. c) $P(M=3)\approx0.1380$.

54748912

A rare-event monitor succeeds independently on each cycle with probability $0.15$. Let $X$ be the cycle of the first success. a) Find the mean of $X$. b) Find the $90$th percentile, defined as the smallest integer $n$ for which $P(X\le n)\ge0.90$. c) Explain why the percentile is much larger than the mean.

Hints

- Treat the long-run average and the percentile as different summaries. - Use a cumulative probability inequality for the percentile. - Connect the comparison to the shape of a waiting-time distribution.

Solution

1. The mean is $E(X)=\frac{1}{0.15}\approx6.67$ cycles. 2. Solve $1-(0.85)^n\ge0.90$, which gives $n\ge\frac{\log(0.10)}{\log(0.85)}\approx14.17$. 3. The smallest integer is $n=15$, and $P(X\le15)=1-(0.85)^{15}\approx0.9126$. 4. The geometric distribution is right-skewed, so a relatively small proportion of long waits extends the upper percentiles far beyond the mean.

Answer

a) $E(X)\approx6.67$ cycles. b) The $90$th percentile is $15$ cycles. c) The long right tail makes high percentiles substantially larger than the mean.

54749312

For a geometric random variable $X$ that counts trials through the first success, $\frac{P(X>10)}{P(X>6)}=0.30$. Find the trial success probability $p$.

Hints

- Express both tail probabilities using the same failure rate. - Simplify the ratio before solving. - Convert the recovered failure rate to the requested success rate.

Solution

1. For a geometric variable, $P(X>k)=(1-p)^k$. 2. Therefore, $\frac{P(X>10)}{P(X>6)}=(1-p)^4=0.30$. 3. Thus, $1-p=0.30^{1/4}$, so $p=1-0.30^{1/4}\approx0.2599$.

Answer

$p=1-0.30^{1/4}\approx0.2599$.

54749612

On each independent trial, a component passes only if it passes both an electrical check and a mechanical check. The checks are independent within a trial, with pass probabilities $0.60$ and $0.70$. Let $X$ be the trial number of the first component that passes both checks. a) Find the geometric success probability for $X$. b) Find $E(X)$. c) Find $P(X>3)$.

Hints

- Define success for one complete trial before using a waiting-time model. - Use the relationship between the two checks within a trial. - Interpret a wait beyond three trials as three combined-trial failures.

Solution

1. A trial succeeds only when both checks pass, so $p=(0.60)(0.70)=0.42$. 2. The mean is $E(X)=\frac{1}{0.42}\approx2.381$ trials. 3. The failure probability is $0.58$, so $P(X>3)=(0.58)^3\approx0.1951$.

Answer

a) $p=0.42$. b) $E(X)\approx2.381$ trials. c) $P(X>3)\approx0.1951$.

54750012

A geometric random variable $X$ counts trials through the first success, where $0<p<1$. Suppose $P(X>7)=P(X=7)$. Find the success probability $p$.

Hints

- Write the tail and point probabilities using the same success rate. - Cancel the common nonzero factor shared by both sides. - Check that the resulting value lies in the stated interval.

Solution

1. The tail probability is $P(X>7)=(1-p)^7$. 2. The point probability is $P(X=7)=(1-p)^6p$. 3. Since $0<p<1$, divide by $(1-p)^6$ to obtain $1-p=p$. 4. Therefore, $p=\frac{1}{2}$.

Answer

$p=\frac{1}{2}$.

54750112

A geometric random variable $X$ counts trials through the first success, with $p=0.30$. Given that the first trial failed, find the conditional mean and standard deviation of $X$.

Hints

- Reframe the process after the known failed trial. - Separate the already-used trial from the remaining random wait. - A fixed shift affects center and spread differently.

Solution

1. Given the first failure, the remaining number of trials through success has the original geometric distribution by memorylessness. 2. Thus, $X\mid X>1$ has the same distribution as $1+X'$, where $X'$ is geometric with $p=0.30$. 3. The conditional mean is $1+\frac{1}{0.30}\approx4.333$. 4. Adding $1$ does not change spread, so the conditional standard deviation is $\frac{\sqrt{0.70}}{0.30}\approx2.789$.

Answer

$E(X\mid X>1)\approx4.333$ trials and $\sigma_{X\mid X>1}\approx2.789$ trials.

54750312

Ten independent devices are each tested repeatedly until their first success. Every test succeeds with probability $0.30$, independently across tests and devices. Let $X_i$ be the test count for device $i$. Find the probability that at least one device requires more than $8$ tests.

Hints

- Start with the tail probability for one device. - Use the complement of the event involving at least one device. - Apply independence across the ten device-level events.

Solution

1. For one device, $P(X_i>8)=(0.70)^8=0.05764801$. 2. The probability that one device succeeds within $8$ tests is $1-(0.70)^8$. 3. Independence across devices gives the probability all ten succeed within $8$ tests as $[1-(0.70)^8]^{10}$. 4. Therefore, the requested probability is $1-[1-(0.70)^8]^{10}\approx0.4478$.

Answer

$1-[1-(0.70)^8]^{10}\approx0.4478$.

54746612

A technician can use either of two independent repeated-search methods until a target is found. An attempt counts toward a time limit only if the attempt is completed within that limit. Method A takes $4$ seconds per attempt and succeeds with probability $0.20$ per attempt. Method B takes $7$ seconds per attempt and succeeds with probability $0.30$ per attempt. a) Compare the expected search times. b) For each method, find the probability of finding the target within $20$ seconds. c) Which method is better by each criterion?

Hints

- Convert an expected number of attempts into expected elapsed time. - Count how many complete attempts fit inside the time limit for each method. - Use the chance of no success as the complement for the deadline comparison.

Solution

1. Method A has expected time $4\left(\frac{1}{0.20}\right)=20$ seconds. 2. Method B has expected time $7\left(\frac{1}{0.30}\right)\approx23.33$ seconds. 3. Within $20$ seconds, Method A completes $5$ attempts, so its success probability is $1-(0.80)^5\approx0.6723$. 4. Within $20$ seconds, Method B completes $2$ attempts, so its success probability is $1-(0.70)^2=0.51$. 5. Method A is better under both criteria.

Answer

a) Method A: $20$ seconds; Method B: approximately $23.33$ seconds. b) Method A: approximately $0.6723$; Method B: $0.51$. c) Method A is better by both expected time and probability of success within $20$ seconds.

54747212

A geometric random variable $X$ counts trials through the first success and has variance $6$. Find the success probability $p$ and the mean of $X$.

Hints

- Relate the given spread measure to the unknown trial success rate. - Expect an algebraic equation with more than one root. - Check candidate roots against the constraints on a probability.

Solution

1. For a geometric variable, $\operatorname{Var}(X)=\frac{1-p}{p^2}$. 2. Set $\frac{1-p}{p^2}=6$, giving $6p^2+p-1=0$. 3. The roots are $p=\frac{1}{3}$ and $p=-\frac{1}{2}$; only $p=\frac{1}{3}$ is a valid probability. 4. The mean is $E(X)=\frac{1}{p}=3$ trials.

Answer

$p=\frac{1}{3}$, and $E(X)=3$ trials.

54747912

A laboratory allows at most $4$ attempts to obtain a valid reading. Each attempt succeeds independently with probability $0.25$. Let $N$ be the number of attempts actually used; if the first three attempts fail, then $N=4$ whether or not the fourth succeeds. a) Construct the probability distribution of $N$. b) Find $E(N)$.

Hints

- Separate stopping early from reaching the final allowed attempt. - Notice that the last probability includes two possible outcomes on the last attempt. - Use the completed distribution to compute the long-run average attempts used.

Solution

1. For $N=1,2,3$, the process stops with the first success, giving probabilities $0.25$, $(0.75)(0.25)=0.1875$, and $(0.75)^2(0.25)=0.140625$. 2. The event $N=4$ occurs whenever the first three attempts fail, so $P(N=4)=(0.75)^3=0.421875$. 3. The expected value is $E(N)=1(0.25)+2(0.1875)+3(0.140625)+4(0.421875)=\frac{175}{64}\approx2.734$.

Answer

a) <table> <tr><th>$n$</th><th>$1$</th><th>$2$</th><th>$3$</th><th>$4$</th></tr> <tr><th>$P(N=n)$</th><td>$0.25$</td><td>$0.1875$</td><td>$0.140625$</td><td>$0.421875$</td></tr> </table> b) $E(N)=\frac{175}{64}\approx2.734$ attempts.

54748212

A geometric random variable $X$ counts attempts through the first success, with success probability $0.30$. Given that success occurs within the first $5$ attempts, find the conditional expected attempt number $E(X\mid X\le5)$.

Hints

- Restrict the possible waiting times to those allowed by the condition. - Renormalize their probabilities before taking a weighted average. - Keep the conditioning probability separate from the numerator of the conditional mean.

Solution

1. The conditioning probability is $P(X\le5)=1-(0.70)^5=0.83193$. 2. The weighted probability sum over the allowed values is $\sum_{k=1}^{5}k(0.70)^{k-1}(0.30)=1.93275$. 3. Therefore, $E(X\mid X\le5)=\frac{1.93275}{0.83193}\approx2.323$.

Answer

$E(X\mid X\le5)\approx2.323$ attempts.

54748312

A researcher can run up to $3$ independent searches each day. Each search finds the target with probability $0.20$. The researcher stops for the day after a success; if all three searches fail, the process resumes the next day under the same conditions. Let $D$ be the day on which the first success occurs. a) Explain why $D$ has a geometric distribution and find its daily success probability. b) Find $P(D=2)$. c) Find $E(D)$.

Hints

- Treat one full day as a single repeated trial. - Find the chance that a day contains at least one success. - After defining the day-level trial, apply the waiting-time model at that level.

Solution

1. A day is a success if at least one of its three searches succeeds. Different days have identical independent search conditions. 2. The daily success probability is $r=1-(0.80)^3=0.488$, so $D$ is geometric with parameter $0.488$. 3. $P(D=2)=(1-r)r=(0.512)(0.488)=0.249856\approx0.2499$. 4. $E(D)=\frac{1}{0.488}\approx2.049$ days.

Answer

a) $D$ is geometric with daily success probability $0.488$. b) $P(D=2)\approx0.2499$. c) $E(D)\approx2.049$ days.

54748512

A geometric random variable $X$ counts trials through the first success, with $p=0.40$. a) Find the probability that $X$ is odd. b) Find the probability that $X$ is even.

Hints

- Write the first few point probabilities for one parity. - Identify the common multiplier between consecutive terms in that list. - Use the fact that every positive integer is either odd or even.

Solution

1. Odd values are $1,3,5,\ldots$, so their probabilities form $0.40+(0.60)^2(0.40)+(0.60)^4(0.40)+\cdots$. 2. This geometric series has sum $\frac{0.40}{1-(0.60)^2}=\frac{5}{8}=0.625$. 3. The even probability is the complement, $1-\frac{5}{8}=\frac{3}{8}=0.375$.

Answer

a) $P(X\text{ is odd})=\frac{5}{8}=0.625$. b) $P(X\text{ is even})=\frac{3}{8}=0.375$.

54748712

Independent geometric random variables $X$ and $Y$ count trials through first success, with success probabilities $0.20$ and $0.30$, respectively. Let $T=\max(X,Y)$, the number of rounds needed until both processes have succeeded at least once. Find $P(T\le4)$. Is $T$ itself geometric? Explain.

Hints

- Rewrite the maximum event as simultaneous conditions on the two component waits. - Use the independence of the two processes after finding each cumulative probability. - For the model question, examine whether the state of the process is unchanged after every unfinished round.

Solution

1. The event $T\le4$ occurs when both $X\le4$ and $Y\le4$. 2. Independence gives $P(T\le4)=[1-(0.80)^4][1-(0.70)^4]\approx0.4486$. 3. The chance that the process finishes in a round depends on whether neither, one, or both component processes have already succeeded. Therefore, $T$ does not have a constant success probability from round to round and is not geometric.

Answer

$P(T\le4)=[1-(0.80)^4][1-(0.70)^4]\approx0.4486$. The variable $T$ is not geometric.

54749012

A geometric random variable $X$ counts trials through the first success. It is known that $P(X=2\mid X\le2)=0.40$. Find the success probability $p$.

Hints

- Write the exact and cumulative probabilities using the same failure rate. - Simplify the conditional ratio before solving. - Convert the resulting failure probability back to a success probability.

Solution

1. Let $q=1-p$. Then $P(X=2)=qp$ and $P(X\le2)=p+qp=p(1+q)$. 2. The conditional probability is $\frac{q}{1+q}=0.40$. 3. Solving gives $q=\frac{2}{3}$, so $p=1-q=\frac{1}{3}$.

Answer

$p=\frac{1}{3}$.

54749112

Before a sequence of trials begins, a hidden setting is chosen with equal probability. Under the low setting, every trial succeeds with probability $0.20$; under the high setting, every trial succeeds with probability $0.80$. Conditional on the chosen setting, trials are independent. Let $X$ be the trial of the first success. a) Find $P(X=3)$. b) Explain why using a geometric model with the average success probability $0.50$ is incorrect.

Hints

- Condition on each possible setting before combining probabilities. - Keep the setting fixed throughout a sequence of trials. - Compare the resulting mixture with what a single constant-rate model would predict.

Solution

1. Under the low setting, $P(X=3)=(0.80)^2(0.20)=0.128$. 2. Under the high setting, $P(X=3)=(0.20)^2(0.80)=0.032$. 3. Averaging over the equally likely settings gives $P(X=3)=0.5(0.128)+0.5(0.032)=0.080$. 4. A geometric model with $p=0.50$ would give $(0.50)^2(0.50)=0.125$. The shared hidden setting makes trial outcomes dependent when the setting is not observed, so averaging the two success rates does not create a geometric process.

Answer

a) $P(X=3)=0.080$. b) The hidden setting persists across trials, creating dependence in the unconditional process. A geometric model with $p=0.50$ would incorrectly give $0.125$.

54749412

A geometric random variable $X$ counts trials through the first success, with $p=0.25$. Find the probability that the first success occurs on a trial number divisible by $4$.

Hints

- List the trial numbers that meet the divisibility condition. - Write the first few point probabilities and identify their common ratio. - Sum the repeating infinite pattern.

Solution

1. The possible values are $4,8,12,\ldots$. 2. Their probabilities form $(0.75)^3(0.25)+(0.75)^7(0.25)+(0.75)^{11}(0.25)+\cdots$. 3. This is a geometric series with first term $(0.75)^3(0.25)$ and common ratio $(0.75)^4$. 4. The sum is $\frac{(0.75)^3(0.25)}{1-(0.75)^4}=\frac{27}{175}\approx0.1543$.

Answer

$P(X\text{ is divisible by }4)=\frac{27}{175}\approx0.1543$.

54749812

A geometric random variable $X$ counts trials through the first success, with $p=0.30$. Given that $X$ is odd, find $P(X=3\mid X\text{ is odd})$.

Hints

- Write the requested conditional probability as a ratio. - The conditioning event contains infinitely many possible waiting times. - Look for a repeating pattern in the probabilities of the allowed values.

Solution

1. The point probability is $P(X=3)=(0.70)^2(0.30)=0.147$. 2. The probability of an odd waiting time is $0.30+(0.70)^2(0.30)+(0.70)^4(0.30)+\cdots=\frac{0.30}{1-(0.70)^2}=\frac{10}{17}$. 3. Therefore, $P(X=3\mid X\text{ is odd})=\frac{0.147}{10/17}=0.2499$.

Answer

$P(X=3\mid X\text{ is odd})=0.2499$.

54750212

A geometric random variable $X$ counts trials through the first success, where $0<p<1$. Suppose $P(X\le3)=2P(X=1)$. Find $p$.

Hints

- Express the cumulative and first-trial probabilities using the failure rate. - Look for a factorization before solving the resulting equation. - Retain only a root that represents a probability.

Solution

1. Let $q=1-p$. The condition becomes $1-q^3=2p=2(1-q)$. 2. Factor $1-q^3=(1-q)(1+q+q^2)$. Since $q\ne1$, cancel $1-q$ to get $1+q+q^2=2$. 3. Thus, $q^2+q-1=0$, and the valid root is $q=\frac{\sqrt{5}-1}{2}$. 4. Therefore, $p=1-q=\frac{3-\sqrt{5}}{2}\approx0.3820$.

Answer

$p=\frac{3-\sqrt{5}}{2}\approx0.3820$.

54750412

For a geometric random variable $X$ that counts trials through the first success, the ratio of the standard deviation to the mean is $0.80$. Find the success probability $p$ and the mean of $X$.

Hints

- Form the requested ratio before substituting the given value. - Simplify common factors in the two distribution summaries. - Convert the resulting failure probability to a success probability.

Solution

1. For a geometric variable, $\frac{\sigma_X}{E(X)}=\frac{\sqrt{1-p}/p}{1/p}=\sqrt{1-p}$. 2. Thus, $\sqrt{1-p}=0.80$, so $1-p=0.64$ and $p=0.36$. 3. The mean is $E(X)=\frac{1}{0.36}\approx2.7778$ trials.

Answer

$p=0.36$, and $E(X)\approx2.7778$ trials.

54750512

A system succeeds independently with probability $0.25$ on each attempt. Before testing begins, an attempt limit $L$ is chosen uniformly from $1,2,3,4,5$, independently of all test outcomes. Testing stops at the first success or when the limit is reached. Find the probability that the system succeeds before testing stops.

Hints

- Condition on each possible attempt limit. - For a fixed limit, use the complement of failing every allowed attempt. - Combine the conditional probabilities using the distribution of the limit.

Solution

1. Conditional on $L=\ell$, the probability of success by the limit is $1-(0.75)^\ell$. 2. Average these five conditional probabilities because the limits are equally likely: $\frac{1}{5}\sum_{\ell=1}^{5}[1-(0.75)^\ell]$. 3. The value is $\frac{1}{5}(0.25+0.4375+0.578125+0.68359375+0.7626953125)\approx0.5424$.

Answer

The probability of success before stopping is approximately $0.5424$.

54749712

A decoding process succeeds independently with probability $0.50$ on each attempt. Attempt $1$ takes $1$ second, attempt $2$ takes $2$ seconds, and in general attempt $k$ takes $k$ seconds. Let $X$ be the attempt of the first success and $T$ the total elapsed time through that attempt. Find $E(T)$.

Hints

- Express the accumulated time in terms of the random attempt number. - Rewrite the expected value using moments of the waiting-time variable. - Use the relationship between variance, the mean, and the second moment.

Solution

1. If success occurs on attempt $X$, then $T=1+2+\cdots+X=\frac{X(X+1)}{2}$. 2. For $p=0.50$, $E(X)=2$ and $\operatorname{Var}(X)=\frac{0.50}{0.50^2}=2$. 3. Therefore, $E(X^2)=\operatorname{Var}(X)+[E(X)]^2=2+4=6$. 4. Thus, $E(T)=\frac{E(X^2)+E(X)}{2}=\frac{6+2}{2}=4$ seconds.

Answer

$E(T)=4$ seconds.

54749912

Two independent processes begin at the same time and each is attempted once per round. Process A succeeds with probability $0.40$ per round, and process B succeeds with probability $0.25$ per round. Let $X$ and $Y$ be their first-success rounds. Find $P(X<Y)$, $P(X=Y)$, and $P(Y<X)$.

Hints

- Classify a round as no decision, A first, a tie, or B first. - Earlier rounds may repeat the no-decision outcome any number of times. - Check that the three final comparison probabilities sum to one.

Solution

1. A round in which neither process succeeds has probability $(0.60)(0.75)=0.45$. 2. A first decisive round favors A when A succeeds and B fails, with probability $(0.40)(0.75)=0.30$. Summing over any number of earlier double failures gives $P(X<Y)=\frac{0.30}{1-0.45}=\frac{6}{11}$. 3. A tie on the first decisive round has probability $(0.40)(0.25)=0.10$, so $P(X=Y)=\frac{0.10}{0.55}=\frac{2}{11}$. 4. B wins with decisive-round probability $(0.60)(0.25)=0.15$, so $P(Y<X)=\frac{0.15}{0.55}=\frac{3}{11}$.

Answer

$P(X<Y)=\frac{6}{11}$, $P(X=Y)=\frac{2}{11}$, and $P(Y<X)=\frac{3}{11}$.

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