Business Statistics Quiz: Poisson Processes

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Poisson ProcessesQuestion 1 of 17

A telecommunications network experiences packet losses according to a Poisson process. In a test period, 3 losses occur in the first 10 minutes, and 7 losses occur in the subsequent 20 minutes. Using maximum likelihood estimation, what is the estimated probability that exactly 2 losses will occur in the next 5-minute interval?

e^{-\frac{1}{3}} \cdot \frac{(\frac{1}{3})^2}{2!} \approx 0.0613

e^{-\frac{5}{6}} \cdot \frac{(\frac{5}{6})^2}{2!} \approx 0.1912

e^{-\frac{5}{3}} \cdot \frac{(\frac{5}{3})^2}{2!} \approx 0.2572

e^{-2} \cdot \frac{2^2}{2!} \approx 0.2707

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Business Statistics Quiz

Business Statistics Quiz: Poisson Processes

Practice Poisson Processes in Business Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Poisson Processes, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$e^{-\frac{1}{3}} \cdot \frac{(\frac{1}{3})^2}{2!} \approx 0.0613$
$e^{-\frac{5}{6}} \cdot \frac{(\frac{5}{6})^2}{2!} \approx 0.1912$
$e^{-\frac{5}{3}} \cdot \frac{(\frac{5}{3})^2}{2!} \approx 0.2572$ (correct answer)
$e^{-2} \cdot \frac{2^2}{2!} \approx 0.2707$

Explanation: The MLE for a Poisson process rate is total events divided by total time. Total losses = 3 + 7 = 10, total time = 30 minutes. Estimated rate = 10/30 = 1/3 losses per minute. For 5 minutes, expected losses = 5 × (1/3) = 5/3. P(X = 2) = e^(-5/3) × (5/3)²/2! ≈ 0.2572. Choice A uses rate 1/3 for 5 minutes incorrectly as 1/3 total. Choice B uses rate 1/6 per minute. Choice D uses rate 2/5 per minute.

Question 2

Machine failures in a manufacturing plant follow a Poisson process with an average of 2.4 failures per 8-hour shift. If the plant operates three shifts per day, and failures in different shifts are independent, what is the probability that the total number of failures in a day is exactly 6?

$e^{-2.4} \cdot \frac{2.4^6}{6!} \approx 0.0241$
$e^{-7.2} \cdot \frac{7.2^6}{6!} \approx 0.1490$ (correct answer)
$3 \times e^{-2.4} \cdot \frac{2.4^6}{6!} \approx 0.0723$
$\left(e^{-2.4} \cdot \frac{2.4^2}{2!}\right)^3 \approx 0.2613$

Explanation: For three 8-hour shifts, the total rate is 3 × 2.4 = 7.2 failures per day. Since we want exactly 6 failures total in a day, we use the Poisson formula with λ = 7.2: P(X = 6) = e^(-7.2) × 7.2⁶/6! ≈ 0.1490. Choice A uses only one shift's rate. Choice C incorrectly multiplies by 3, treating shifts as separate events rather than combining rates. Choice D assumes exactly 2 failures per shift, which is not what the question asks.

Question 3

An e-commerce company's servers handle purchase orders from two separate, independent marketing campaigns. Orders from campaign A arrive at a rate of 3 per hour. Orders from campaign B arrive at a rate of 5 per hour. Both are Poisson processes. What is the probability that the company receives fewer than 12 total orders during a two-hour period?

0.084
0.155 (correct answer)
0.267
0.394

Explanation: The sum of independent Poisson processes is also a Poisson process.\n1. The total arrival rate is the sum of the individual rates:

\lambda_{total} = \lambda_A + \lambda_B = 3 + 5 = 8

orders per hour.\n2. The time period is 2 hours. So, the mean number of orders for this period is

\mu = \lambda_{total} \times t = 8 \times 2 = 16

.\n3. We need to find the probability of receiving fewer than 12 total orders, which is

P(X < 12) = P(X \le 11)

.\n4. We must sum the Poisson probabilities for k = 0, 1, ..., 11, with a mean of

\mu=16

. Using a calculator or statistical software for the Poisson cumulative distribution function:

P(X \le 11) \approx 0.155

.\nDistractor A is

P(X=12)

. Distractor C is

P(X \le 13)

. Distractor D is

P(X \le 14)

, a common off-by-one error (calculating for 'up to 12' or 'less than or equal to 12').

Question 4

A city's emergency response system models incoming calls using a Poisson process. Which of the following scenarios would most likely violate the assumptions of the Poisson process model for a full 24-hour day?

The number of calls received during the first hour of the day is independent of the number of calls received during the last hour.
The probability of two calls arriving at the exact same millisecond is effectively zero.
The average number of calls received per hour is significantly higher during evening rush hour than in the middle of the night. (correct answer)
The time between any two consecutive calls follows an exponential distribution.

Explanation: A key assumption of a simple Poisson process is that the rate

\lambda

is constant over time. Scenario C describes a situation where the rate of calls changes depending on the time of day (non-homogeneous Poisson process). This violates the constant-rate assumption of the basic Poisson model. The other options describe properties that are consistent with a Poisson process: (A) Independence of non-overlapping intervals, (B) No simultaneous events, and (D) Exponentially distributed inter-arrival times. While non-homogeneous Poisson processes exist, this scenario violates the assumptions of the standard, constant-rate model typically first studied.

Question 5

A manufacturer finds that imperfections in a copper wire occur at a rate of 0.2 imperfections per foot. The wire is sold in spools. The manufacturer wants 99% of spools to have at least one imperfection. What is the minimum length of wire, in feet, that should be on each spool to meet this requirement?

23 feet (correct answer)
12 feet
3 feet
45 feet

Explanation: When you encounter a problem about imperfections occurring at a constant rate, you're dealing with a Poisson distribution. This distribution models events that happen randomly over time, space, or in this case, length. Given that imperfections occur at 0.2 per foot, we need to find the minimum wire length where 99% of spools have at least one imperfection. This means only 1% should have zero imperfections. For a Poisson distribution, the probability of zero events is

P(X = 0) = e^{-\lambda}

, where

\lambda

is the expected number of imperfections (rate × length). We want

P(X = 0) = 0.01

, so:

e^{-\lambda} = 0.01

-\lambda = \ln(0.01) = -4.605

\lambda = 4.605

Since

\lambda = 0.2 \times \text{length}

\text{length} = \frac{4.605}{0.2} = 23.025

feet Therefore, the minimum length is 23 feet. Let's examine why the other options fail: Option B (12 feet) gives

\lambda = 2.4

, resulting in

P(X = 0) = e^{-2.4} = 0.091

or 9.1% with zero imperfections—far above the required 1%. Option C (3 feet) gives

\lambda = 0.6

and

P(X = 0) = 0.549

, meaning over half the spools would have no imperfections. Option D (45 feet) would work but exceeds the minimum requirement, making it inefficient. Remember: in Poisson problems asking for "at least one," always work with the complement—finding when zero events occur, then subtracting from 1.

Question 6

The number of trades executed per minute on a stock exchange follows a Poisson distribution with a mean of $\lambda=9.5$ . The mode of a Poisson distribution is the integer value $k$ that has the highest probability. For $\lambda=9.5$ , the probabilities $P(X=k)$ increase until the mode and then decrease. What is the largest integer $k$ for which the probability of $k$ trades is less than the probability of $k+1$ trades?

7
8 (correct answer)
9
10

Explanation: This question tests the understanding of the shape of the Poisson probability mass function.\n1. The relationship between consecutive Poisson probabilities is

P(X=k+1) = P(X=k) \times \frac{\lambda}{k+1}

.\n2. We are looking for the largest integer

k

such that

P(X=k) < P(X=k+1)

.\n3. This inequality is equivalent to

1 < \frac{P(X=k+1)}{P(X=k)}

, which simplifies to

1 < \frac{\lambda}{k+1}

.\n4. Rearranging the inequality gives

k+1 < \lambda

, or

k < \lambda - 1

.\n5. Given

\lambda = 9.5

, we need to find the largest integer

k

such that

k < 9.5 - 1

, which is

k < 8.5

.\n6. The largest integer

k

that satisfies this condition is

k=8

.\nWhen

k=8

k+1=9

, which is less than

\lambda=9.5

, so

P(X=8) < P(X=9)

. When

k=9

k+1=10

, which is greater than

\lambda=9.5

, so

P(X=9) > P(X=10)

. The mode is 9. Therefore, the largest k for which P(k) is less than P(k+1) is 8. Distractor C, 9, is the mode itself, where the probability is highest. Distractor A, 7, also satisfies the condition, but it is not the largest such integer.

Question 7

A critical server component has a failure rate that follows a Poisson process, with a mean of one failure per 4,000 hours of operation. Technicians perform a check and find the component is working. Given that the component has already operated for 1,000 hours without failure since the last replacement, what is the probability that it will operate for at least another 500 hours without failure?

0.125
0.779
0.882 (correct answer)
0.975

Explanation: This question tests the memoryless property of the exponential distribution, which describes the waiting time for the next event in a Poisson process.

The memoryless property states that the probability of a future event is independent of the past. The fact that the component has already operated for 1,000 hours without failure is irrelevant to its future performance.
The failure rate is $\lambda = 1 / 4000$ failures per hour.\n3. We need to find the probability that the component operates for at least another 500 hours. Let T be the time until the next failure. We need $P(T > 500)$ .\n4. Using the survival function of the exponential distribution: $P(T > t) = e^{-\lambda t}$ .\n5. $P(T > 500) = e^{-(1/4000) \times 500} = e^{-500/4000} = e^{-1/8} = e^{-0.125} \approx 0.882$ .\nDistractor B, 0.779, is the probability of surviving 1,000 hours ( $e^{-1000/4000} = e^{-0.25}$ ), which ignores the question about the next 500 hours. Distractor A, 0.125, is the value of the exponent, not the probability. Distractor D might arise from miscalculating the rate.

Question 8

A machine fills bags with coffee. The machine processes orders at a rate of 1 order every 3 minutes, following a Poisson process. What is the probability that it takes the machine more than 15 minutes to process the fourth order?

0.125
0.264 (correct answer)
0.353
0.647

Explanation: This question about the time until the k-th event can be transformed into a question about the number of events in a fixed time interval.

The statement 'it takes more than 15 minutes to process the fourth order' is logically equivalent to 'in the first 15 minutes, fewer than 4 orders were processed'.
'Fewer than 4 orders' means $X \le 3$ orders.\n3. First, determine the rate of the Poisson process. The rate is 1 order per 3 minutes, which is $\lambda = 1/3$ orders per minute.\n4. The time interval is $t=15$ minutes.\n5. The mean number of orders in this interval is $\mu = \lambda \times t = (1/3) \times 15 = 5$ .\n6. Now, we need to calculate $P(X \le 3)$ for a Poisson distribution with a mean of $\mu=5$ .\n7. $P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$ . Using a Poisson CDF calculator with $\mu=5$ , we find $P(X \le 3) \approx 0.265$ . This matches choice B.\nDistractor D, 0.647, would be closer to $P(X>4)$ , which is a misinterpretation of the question. Distractor C, 0.353, might be $P(X=4)$ or another miscalculation. Distractor A is $P(X\le2)$ .

Question 9

A traffic analyst is modeling the arrival of cars at a rural intersection and considers using a Poisson process. The analyst collects data for 100 separate 5-minute intervals. The sample mean number of arrivals per interval was found to be 3.8, and the sample variance was 7.9. Based on these statistics, what is the most appropriate conclusion?

The Poisson model is likely inappropriate because the sample variance is substantially different from the sample mean. (correct answer)
The Poisson model is likely inappropriate because the sample mean is not an integer.
The Poisson model is appropriate, as the mean arrival rate is constant across intervals.
The Poisson model is appropriate, as the total number of cars observed (380) is large enough for the model to apply.

Explanation: When evaluating whether data fits a Poisson process, you need to check if the key assumptions are met. The most fundamental property of a Poisson distribution is that its variance equals its mean (

\sigma^2 = \mu

). This equality is a mathematical requirement, not just a rough approximation. In this problem, the sample mean is 3.8 arrivals per interval, but the sample variance is 7.9 - more than double the mean. This substantial difference (variance is 208% of the mean) violates the core assumption of the Poisson model, suggesting the data exhibits overdispersion. This could indicate clustering of arrivals, non-constant arrival rates, or other factors that make Poisson inappropriate. Looking at the wrong answers: Option B incorrectly assumes the sample mean must be an integer - while individual counts are integers, the sample mean of those counts can be any value. Option C makes an unsupported claim about constant arrival rates without evidence, and this property alone wouldn't validate the model anyway. Option D misunderstands sample size requirements - having 380 total observations doesn't override the fundamental distributional mismatch between mean and variance. The correct answer is A because the large discrepancy between sample variance and sample mean indicates the Poisson model assumptions are violated. Study tip: Always check if variance approximately equals the mean when evaluating Poisson models. If variance is much larger (overdispersion) or smaller (underdispersion), consider alternative models like negative binomial or look for underlying causes in the data collection process.

Question 10

A small business's IT system experiences security alerts at a Poisson rate of 0.8 per day. Each alert costs the company $200 to investigate. The business is considering a security service contract that costs $120 per day. From a purely financial perspective, not buying the contract is the better decision if the daily investigation cost is less than the contract cost. What is the probability that not buying the contract is the correct decision on a given day?

0.359
0.449 (correct answer)
0.809
0.953

Explanation: This problem requires translating a business decision into a probability calculation.\n1. Let X be the number of security alerts on a given day.

X

follows a Poisson distribution with

\lambda = 0.8

.\n2. The daily investigation cost is

200 \times X

. The contract cost is $120.\n3. Not buying the contract is the correct decision if

200X < 120

.\n4. Solving for X:

X < 120 / 200

, which means

X < 0.6

.\n5. Since X must be a non-negative integer, the only value that satisfies this condition is

X=0

.\n6. Therefore, the question is asking for the probability that

X=0

.\n7. Using the Poisson formula:

P(X=0) = \frac{e^{-0.8}0.8^0}{0!} = e^{-0.8} \approx 0.449

.\nDistractor C calculates

P(X \le 1) = P(X=0) + P(X=1) = e^{-0.8}(1+0.8) = 0.449 \times 1.8 = 0.809

, incorrectly including the case of one alert. The cost for one alert is $200, which is greater than the $120 contract. Distractor A is

P(X=1)

. Distractor D is

P(X \le 2)

Question 11

A busy intersection sees accidents occurring as a Poisson process at a rate of 2.5 per month. What is the probability that the third accident of the year occurs after the end of January (i.e., after the first month)?

0.456
0.544 (correct answer)
0.677
0.891

Explanation: This question relates the waiting time for the k-th event (a Gamma distribution) to the count of events in an interval (a Poisson distribution).

The event 'the third accident occurs after the first month' is equivalent to the event 'at most 2 accidents occur in the first month'.
The rate of accidents is $\lambda = 2.5$ per month.\n3. We need to calculate $P(X \le 2)$ for a Poisson distribution with mean $\lambda = 2.5$ .\n4. $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ .\n5. Using the Poisson formula $P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$ :\n $P(X=0) = e^{-2.5} \approx 0.0821$ \n $P(X=1) = e^{-2.5} \frac{2.5^1}{1!} \approx 0.2052$ \n $P(X=2) = e^{-2.5} \frac{2.5^2}{2!} \approx 0.2565$ \n6. $P(X \le 2) = 0.0821 + 0.2052 + 0.2565 = 0.5438 \approx 0.544$ .\nDistractor A is $1 - 0.544 = 0.456$ , which is the probability the third accident occurs within the first month. Distractor C is $P(X \le 3)$ , an off-by-one error. Distractor D might result from using an annual rate and trying to work with a Gamma distribution directly, which is more complex and prone to error.

Question 12

A website receives visitors according to a Poisson process at a rate of 12 per hour. A monitoring system is set up to send an alert if no visitors arrive for a certain period of time. What is the minimum monitoring period (in minutes) that must pass with no visitors to ensure the probability of such a quiet period is 5% or less?

15 minutes (correct answer)
10 minutes
20 minutes
25 minutes

Explanation: When you encounter a Poisson process problem asking about the probability of "no events" in a given time period, you're dealing with the exponential distribution. In a Poisson process with rate λ events per unit time, the time between events follows an exponential distribution, and the probability of no events occurring in time t is

P(\text{no events in time } t) = e^{-\lambda t}

. Here, visitors arrive at rate 12 per hour. Converting to minutes: λ = 12/60 = 0.2 visitors per minute. You need the minimum time t where

P(\text{no visitors in time } t) \leq 0.05

. Setting up the equation:

e^{-0.2t} \leq 0.05

Taking the natural logarithm:

-0.2t \leq \ln(0.05) = -2.996

Solving:

t \geq \frac{2.996}{0.2} = 14.98

minutes Since we need the minimum whole number of minutes, this rounds up to 15 minutes, making A correct. Let's check the other options: B (10 minutes) gives

e^{-0.2(10)} = e^{-2} = 0.135

, which is 13.5% - too high. C (20 minutes) gives

e^{-4} = 0.018

or 1.8%, and D (25 minutes) gives

e^{-5} = 0.007

or 0.7%. While C and D satisfy the 5% threshold, they're not the minimum required. Remember: in "minimum time" problems with Poisson processes, solve the exponential inequality exactly, then round up to ensure you meet the threshold requirement. Don't just test the given options.

Question 13

A manufacturing robot fails according to a Poisson process with a mean time between failures of 500 hours. The robot has just been serviced and is running perfectly. What is the probability that it will run for at least 200 hours without a failure?

0.330
0.400
0.670 (correct answer)
0.918

Explanation: The time between events in a Poisson process follows an exponential distribution.\n1. The rate of failure,

\lambda

, is the reciprocal of the mean time between failures.

\lambda = 1 / 500

failures per hour.\n2. The probability that the time to the next failure, T, is greater than some time t is given by

P(T > t) = e^{-\lambda t}

.\n3. We want to find the probability that the robot runs for at least 200 hours, so

t = 200

.\n4.

P(T > 200) = e^{-(1/500) \times 200} = e^{-200/500} = e^{-0.4} \approx 0.670

.\nDistractor D is

e^{-0.085}

, possibly from an error in calculating lambda. Distractor A is

1 - 0.670 = 0.330

, which is the probability it fails within 200 hours. Distractor B is the rate

200/500 = 0.4

, not the probability.

Question 14

A delivery service receives orders according to a Poisson process with rate λ per hour. During peak hours, they observe that the probability of receiving exactly k orders in 1 hour equals the probability of receiving exactly k+1 orders in 1 hour when k = 2. What is the probability that they receive exactly 1 order in a 20-minute period during peak hours?

$e^{-3} \cdot 3 \approx 0.149$
$e^{-\frac{2}{3}} \cdot \frac{2}{3} \approx 0.342$
$e^{-2} \cdot 2 \approx 0.271$
$e^{-1} \cdot 1 \approx 0.368$ (correct answer)

Explanation: When you encounter a Poisson process problem, remember that the key is using the probability mass function

P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

and adjusting the rate parameter for different time periods. First, you need to find the rate λ during peak hours. Given that

P(X = 2) = P(X = 3)

for a 1-hour period:

\frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^3 e^{-\lambda}}{3!}

Simplifying:

\frac{\lambda^2}{2} = \frac{\lambda^3}{6}

This gives us

3\lambda^2 = \lambda^3

, so

\lambda = 3

orders per hour. Now, for a 20-minute period, you must scale the rate: 20 minutes =

\frac{1}{3}

hour, so the effective rate is

\lambda_{20} = 3 \times \frac{1}{3} = 1

. The probability of exactly 1 order in 20 minutes is:

P(X = 1) = \frac{1^1 e^{-1}}{1!} = e^{-1} \approx 0.368

Answer A uses λ = 3 without time adjustment, giving the probability for a full hour rather than 20 minutes. Answer B incorrectly uses λ = 2/3, perhaps confusing the time fraction with the rate. Answer C uses λ = 2, which doesn't match our calculated peak rate of 3 orders per hour. Study tip: In Poisson problems involving time periods, always remember to scale the rate parameter proportionally. If the given rate is per hour and you need a 20-minute probability, multiply the rate by 20/60 = 1/3.

Question 15

A quality control inspector examines products from two independent production lines. Line A produces defects according to a Poisson process with rate 3 per hour, while Line B has rate 2 per hour. During a 2-hour inspection period, what is the probability that the combined number of defects from both lines is exactly 8?

$e^{-5} \cdot \frac{5^8}{8!} + e^{-5} \cdot \frac{5^8}{8!} \approx 0.0653$
$e^{-10} \cdot \frac{10^8}{8!} \approx 0.1126$ (correct answer)
$\left(e^{-6} \cdot \frac{6^4}{4!}\right) \cdot \left(e^{-4} \cdot \frac{4^4}{4!}\right) \approx 0.0228$
$\sum_{k=0}^{8} \left(e^{-6} \cdot \frac{6^k}{k!}\right) \cdot \left(e^{-4} \cdot \frac{4^{8-k}}{(8-k)!}\right) \approx 0.1126$

Explanation: When combining independent Poisson processes, the rates add. Line A: 3 × 2 = 6 defects expected in 2 hours. Line B: 2 × 2 = 4 defects expected. Combined rate = 6 + 4 = 10 defects in 2 hours. P(X = 8) = e^(-10) × 10⁸/8! ≈ 0.1126. Choice A incorrectly adds probabilities instead of combining processes. Choice C assumes exactly 4 defects from each line. Choice D uses the convolution formula, which gives the same answer but is unnecessarily complex since the sum of Poisson processes is also Poisson.

Question 16

A call center receives customer service calls according to a Poisson process with a rate of 8 calls per hour. During a particularly busy period, the center receives 15 calls in the first hour and 12 calls in the second hour. What is the probability that exactly 3 calls will be received in the next 30-minute period?

$e^{-4} \cdot \frac{4^3}{3!} \approx 0.195$ (correct answer)
$e^{-6.75} \cdot \frac{6.75^3}{3!} \approx 0.169$
$e^{-13.5} \cdot \frac{13.5^3}{3!} \approx 0.003$
$e^{-8} \cdot \frac{8^3}{3!} \approx 0.286$

Explanation: In a Poisson process, the rate parameter is fixed and independent of past observations. The historical data (15 and 12 calls) does not affect future probabilities. For a 30-minute period, λ = 8 × 0.5 = 4 calls. P(X = 3) = e^(-4) × 4³/3! ≈ 0.195. Choice B incorrectly uses the average of past observations (27/2 = 13.5 calls per hour, so 6.75 per half hour). Choice C uses the total past observations as the rate. Choice D incorrectly uses the full hourly rate for a 30-minute period.

Question 17

A security system detects intrusion attempts according to a Poisson process with rate 1.2 per day. The system has a backup that activates only when the primary system fails, which happens independently with probability 0.02 per day. On days when the backup is active, the detection rate increases to 1.8 per day. What is the probability of detecting exactly 3 attempts on a randomly selected day?

$0.98 \cdot e^{-1.2} \cdot \frac{1.2^3}{3!} \cdot 0.02 \cdot e^{-1.8} \cdot \frac{1.8^3}{3!} \approx 0.0002$
$e^{-1.236} \cdot \frac{1.236^3}{3!} \approx 0.0859$
$e^{-1.5} \cdot \frac{1.5^3}{3!} \approx 0.1255$
$0.98 \cdot e^{-1.2} \cdot \frac{1.2^3}{3!} + 0.02 \cdot e^{-1.8} \cdot \frac{1.8^3}{3!} \approx 0.0868$ (correct answer)

Explanation: When you encounter a problem involving multiple scenarios with different rates or probabilities, you're dealing with a mixture distribution. The key insight is using the law of total probability to account for all possible ways the event can occur. Here, there are two mutually exclusive scenarios: the primary system working (probability 0.98) with rate 1.2 detections per day, or the primary system failing (probability 0.02) and the backup activating with rate 1.8 detections per day. Since we want exactly 3 detections regardless of which scenario occurs, we must calculate the probability for each scenario and add them together. For the primary system:

P(\text{3 detections | primary}) = e^{-1.2} \cdot \frac{1.2^3}{3!}

For the backup system:

P(\text{3 detections | backup}) = e^{-1.8} \cdot \frac{1.8^3}{3!}

The total probability is:

0.98 \cdot e^{-1.2} \cdot \frac{1.2^3}{3!} + 0.02 \cdot e^{-1.8} \cdot \frac{1.8^3}{3!}

This matches answer D. Answer A incorrectly multiplies the probabilities instead of adding them, suggesting both events must happen simultaneously. Answer B uses a weighted average of the rates (

1.236 = 0.98 \times 1.2 + 0.02 \times 1.8

), which is mathematically incorrect for Poisson processes. Answer C uses the simple average of rates (

\frac{1.2 + 1.8}{2} = 1.5

), ignoring the probabilities entirely. Study tip: For mixture problems, always use the law of total probability: multiply each scenario's probability by its conditional outcome probability, then sum across all scenarios.