Discrete Math Quiz: Poisson Distribution

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Poisson DistributionQuestion 1 of 9

Two independent Poisson processes occur simultaneously: Process A with rate 2.4 events per minute and Process B with rate 1.6 events per minute. What is the probability that exactly 3 total events occur in the next minute?

\frac{e^{-4} \cdot 4^3}{3!} \approx 0.195

\frac{e^{-2.4} \cdot 2.4^3}{3!} + \frac{e^{-1.6} \cdot 1.6^3}{3!} \approx 0.209

\frac{e^{-2.4} \cdot 2.4^3}{3!} \cdot \frac{e^{-1.6} \cdot 1.6^0}{0!} \approx 0.035

\frac{e^{-4} \cdot (2.4 + 1.6)^3}{3!} \approx 0.195

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Discrete Math Quiz

Discrete Math Quiz: Poisson Distribution

Practice Poisson Distribution in Discrete Math 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 Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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

$\frac{e^{-4} \cdot 4^3}{3!} \approx 0.195$ (correct answer)
$\frac{e^{-2.4} \cdot 2.4^3}{3!} + \frac{e^{-1.6} \cdot 1.6^3}{3!} \approx 0.209$
$\frac{e^{-2.4} \cdot 2.4^3}{3!} \cdot \frac{e^{-1.6} \cdot 1.6^0}{0!} \approx 0.035$
$\frac{e^{-4} \cdot (2.4 + 1.6)^3}{3!} \approx 0.195$

Explanation: When two independent Poisson processes are combined, the result is also Poisson with rate λ = 2.4 + 1.6 = 4. So P(exactly 3 events) = e^(-4) × 4³/3!. Choice B incorrectly adds the individual probabilities instead of combining the processes. Choice C calculates the probability of 3 events from A and 0 from B only. Choice D has the right numerical answer but wrong formula (can't factor out the sum inside the exponential).

Question 2

A radioactive sample emits particles at an average rate of 8 particles per second. What is the minimum time interval needed to ensure that the probability of observing at least one particle emission is greater than 0.95?

$\frac{8}{\ln(20)} \text{ seconds} \approx 2.67 \text{ seconds}$
$\frac{\ln(0.95)}{8} \text{ seconds} \approx -0.006 \text{ seconds}$
$\frac{0.95}{8} \text{ seconds} = 0.119 \text{ seconds}$
$\frac{\ln(20)}{8} \text{ seconds} \approx 0.374 \text{ seconds}$ (correct answer)

Explanation: When you encounter radioactive decay problems, you're dealing with a Poisson process where particle emissions occur randomly but at a known average rate. The key insight is that the time between emissions follows an exponential distribution. For a Poisson process with rate λ = 8 particles per second, the probability of observing zero particles in time interval t is

P(0) = e^{-\lambda t} = e^{-8t}

. Therefore, the probability of observing at least one particle is

P(\geq 1) = 1 - e^{-8t}

. We need this probability to exceed 0.95:

1 - e^{-8t} > 0.95

e^{-8t} < 0.05

-8t < \ln(0.05)

t > \frac{-\ln(0.05)}{8} = \frac{\ln(20)}{8}

Since

\ln(20) \approx 2.996

, we get

t \approx 0.374

seconds, confirming answer D. Looking at the wrong answers: A gives

\frac{8}{\ln(20)}

, which incorrectly puts the rate in the numerator instead of denominator. B yields

\frac{\ln(0.95)}{8}

, which is negative because it uses ln(0.95) instead of ln(20) and forgets the negative sign cancellation. C simply divides the desired probability by the rate without any theoretical justification. Study tip: For exponential/Poisson problems, remember that "at least one event" means you calculate the complement of "zero events." The formula

P(\geq 1) = 1 - e^{-\lambda t}

is your go-to tool for these minimum time calculations.

Question 3

A quality control inspector finds that scratches on manufactured panels follow a Poisson distribution with an average of 2.5 scratches per square meter. If a panel has area 1.8 square meters and is rejected when it has 6 or more scratches, what is the probability the panel is accepted?

$1 - \sum_{k=6}^{\infty} \frac{e^{-4.5} \cdot 4.5^k}{k!} \approx 0.831$
$\sum_{k=0}^{5} \frac{e^{-2.5} \cdot 2.5^k}{k!} \approx 0.958$
$\sum_{k=0}^{5} \frac{e^{-4.5} \cdot 4.5^k}{k!} \approx 0.831$ (correct answer)
$\sum_{k=0}^{5} \frac{e^{-1.8} \cdot 1.8^k}{k!} \approx 0.991$

Explanation: When you encounter a Poisson distribution problem, you need to carefully identify the rate parameter (λ) for the specific scenario. The Poisson distribution models the probability of a given number of events occurring in a fixed interval, with the formula

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

. Here, scratches occur at 2.5 per square meter, but your panel is 1.8 square meters. This means the expected number of scratches for this specific panel is λ = 2.5 × 1.8 = 4.5 scratches total. Since the panel is accepted when it has fewer than 6 scratches (0 through 5), you need

P(X ≤ 5) = \sum_{k=0}^{5} \frac{e^{-4.5} \cdot 4.5^k}{k!}

, which is answer C. Answer A uses the correct rate (4.5) but calculates the complement incorrectly—it gives the probability of rejection, not acceptance. Answer B makes a critical error by using λ = 2.5 instead of scaling up for the panel's actual area of 1.8 square meters. This ignores that larger panels have proportionally more scratches. Answer D uses 1.8 as the rate parameter, confusing the area measurement with the scratch rate—a fundamental misunderstanding of how to calculate λ. Study tip: In Poisson problems, always verify that your rate parameter matches the specific interval or area in question. If the given rate is "per unit" but your scenario involves multiple units, you must scale the rate accordingly before applying the formula.

Question 4

A website receives hits according to a Poisson process with rate 3.6 hits per minute. Given that exactly 12 hits occurred in a 4-minute interval, what is the probability that exactly 3 of these hits occurred in the first minute?

$\binom{12}{3} \left(\frac{3.6}{14.4}\right)^3 \left(\frac{10.8}{14.4}\right)^9 \approx 0.258$
$\frac{e^{-3.6} \cdot 3.6^3}{3!} \approx 0.195$
$\binom{12}{3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^9 \approx 0.258$ (correct answer)
$\frac{\binom{12}{3} \cdot 3.6^3 \cdot 10.8^9}{14.4^{12}} \approx 0.085$

Explanation: When you encounter a Poisson process problem with conditional probability, you need to recognize that given a fixed number of events in a larger interval, the distribution of events within subintervals follows a different pattern than the original Poisson distribution. This is a classic application of the conditional uniformity property of Poisson processes. Given that exactly 12 hits occurred in the 4-minute interval, each hit is equally likely to have occurred at any point during those 4 minutes. This transforms the problem into a binomial distribution where each of the 12 hits has probability

\frac{1}{4}

of occurring in the first minute (since 1 minute out of 4 total minutes). The correct approach uses the binomial formula:

\binom{12}{3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^9

, which makes C correct. We're choosing 3 hits out of 12 to occur in the first minute, each with probability

\frac{1}{4}

, while the remaining 9 hits occur in the other 3 minutes with probability

\frac{3}{4}

. A incorrectly uses the rate-based probabilities

\frac{3.6}{14.4}

and

\frac{10.8}{14.4}

, which would apply if we weren't given the condition about exactly 12 hits. B calculates the unconditional probability of exactly 3 hits in the first minute, ignoring the given information entirely. D attempts a complex calculation mixing rates and combinations but doesn't properly apply the conditional uniformity principle. Study tip: In conditional Poisson problems, once you're told the total number of events in an interval, treat each event as equally likely to occur anywhere within that interval—this converts your Poisson problem into a simpler binomial one.

Question 5

A call center receives an average of 4.5 customer complaints per hour during peak times. During a particular 2-hour peak period, what is the probability that they receive exactly 8 complaints, assuming complaints follow a Poisson distribution?

$\frac{e^{-9} \cdot 9^8}{8!} \approx 0.132$ (correct answer)
$\frac{e^{-4.5} \cdot 4.5^8}{8!} \approx 0.066$
$\frac{e^{-8} \cdot 8^8}{8!} \approx 0.140$
$\frac{e^{-2} \cdot 2^8}{8!} \approx 0.001$

Explanation: For a 2-hour period, the expected number of complaints is λ = 4.5 × 2 = 9. Using the Poisson formula P(X = k) = (e^(-λ) × λ^k)/k!, we get P(X = 8) = (e^(-9) × 9^8)/8!. Choice B incorrectly uses the hourly rate instead of adjusting for the 2-hour period. Choice C uses λ = 8 (confusing the desired outcome with the parameter). Choice D uses λ = 2 (misunderstanding the time scaling).

Question 6

A manufacturing process produces defective items at a rate of 0.3 defects per unit produced. If 15 units are produced, what is the probability that there are at most 2 defective items, assuming defects follow a Poisson distribution?

$e^{-4.5}(1 + 4.5 + \frac{4.5^2}{2}) \approx 0.174$ (correct answer)
$e^{-0.3}(1 + 0.3 + \frac{0.3^2}{2}) \approx 0.996$
$1 - e^{-4.5}(1 + 4.5 + \frac{4.5^2}{2}) \approx 0.826$
$e^{-15}(1 + 15 + \frac{15^2}{2}) \approx 0.004$

Explanation: The expected number of defective items is λ = 0.3 × 15 = 4.5. P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = e^(-4.5)[1 + 4.5 + 4.5²/2]. Choice B uses λ = 0.3 without scaling for 15 units. Choice C calculates the complement P(X > 2). Choice D incorrectly uses λ = 15 (confusing the number of units with the defect parameter).

Question 7

Network packets arrive at a router according to two independent Poisson processes: high-priority packets at rate 1.5 per second and low-priority packets at rate 2.8 per second. What is the probability that among the first 3 packets to arrive, exactly 2 are high-priority?

$\frac{e^{-1.5} \cdot 1.5^2}{2!} \cdot \frac{e^{-2.8} \cdot 2.8^1}{1!} \approx 0.030$
$\binom{3}{2} \left(\frac{1.5}{4.3}\right)^2 \left(\frac{2.8}{4.3}\right)^1 \approx 0.337$ (correct answer)
$\binom{3}{2} \left(\frac{1.5}{2.8}\right)^2 \left(\frac{2.8}{1.5}\right)^1 \approx 1.09$
$\frac{1.5^2 \cdot 2.8}{4.3^3} \approx 0.084$

Explanation: When you encounter independent Poisson processes that merge together, the key insight is that the combined process is also Poisson with rate equal to the sum of individual rates. However, this problem asks about the composition of a fixed number of arrivals, which transforms it into a binomial probability question. The total arrival rate is 1.5 + 2.8 = 4.3 packets per second. Given that a packet arrives, the probability it's high-priority is

\frac{1.5}{4.3}

and low-priority is

\frac{2.8}{4.3}

. Among any fixed number of packets, the number of high-priority packets follows a binomial distribution with these probabilities. For exactly 2 high-priority packets out of 3 total, we use the binomial formula:

\binom{3}{2} \left(\frac{1.5}{4.3}\right)^2 \left(\frac{2.8}{4.3}\right)^1 \approx 0.337

. This is answer B. Answer A incorrectly applies the Poisson probability formula directly, treating this as if we're asking for specific counts in each process independently over a time interval. Answer C uses the wrong probability ratio

\frac{1.5}{2.8}

instead of normalizing by the total rate, leading to probabilities that don't sum to 1 and a nonsensical result greater than 1. Answer D omits the binomial coefficient

\binom{3}{2} = 3

, ignoring that there are multiple ways to arrange 2 high-priority packets among 3 total. Remember: when Poisson processes merge and you're asked about composition within a fixed count, convert to binomial probabilities using each process's rate divided by the total rate.

Question 8

A server experiences crashes at a rate of 1.2 crashes per day. If the server has been running without a crash for the past 2 days, what is the probability it will experience exactly 1 crash in the next day?

$\frac{1.2 \cdot e^{-1.2}}{e^{-2.4}} \approx 4.02$
$\frac{1.2 \cdot e^{-1.2}}{1 - e^{-1.2}(1 + 1.2)} \approx 0.540$
$1.2 \cdot e^{-3.6} \approx 0.033$
$1.2 \cdot e^{-1.2} \approx 0.361$ (correct answer)

Explanation: When you encounter questions about random events happening at a constant rate over time, you're dealing with Poisson processes. The key insight here is recognizing the memoryless property - the fact that the server hasn't crashed for 2 days doesn't change the probability of future crashes. In Poisson processes, past events don't influence future probabilities. The server's crash-free period is irrelevant to calculating tomorrow's crash probability. You simply need the probability of exactly 1 crash in the next day, given a rate of 1.2 crashes per day. The Poisson probability formula is:

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

Where λ = 1.2 (rate per day), k = 1 (desired number of crashes), and the time period is 1 day.

P(X = 1) = \frac{1.2^1 e^{-1.2}}{1!} = 1.2 \cdot e^{-1.2} \approx 0.361

This confirms answer D is correct. A incorrectly attempts to condition on the crash-free period by dividing by

e^{-2.4}

, but this misunderstands the memoryless property. B tries to use conditional probability by dividing by the complement of having 0 or 1 crashes, which overcomplicates the problem. C uses λ = 3.6 (combining all three days), showing confusion about the time period we're analyzing. Study tip: In Poisson problems, always identify the relevant time period and rate first. Don't let information about past events distract you - Poisson processes are memoryless.

Question 9

Emergency calls to a fire station follow a Poisson distribution with rate 0.8 calls per hour. What is the probability that the time between two consecutive calls exceeds 2 hours?

$1 - e^{-1.6} \approx 0.798$
$e^{-1.6} \approx 0.202$ (correct answer)
$e^{-0.8} \approx 0.449$
$0.8 \cdot 2 \cdot e^{-1.6} \approx 0.323$

Explanation: When you encounter a Poisson process problem asking about time between events, you need to recognize that while arrivals follow a Poisson distribution, the waiting times between arrivals follow an exponential distribution. This is a fundamental relationship in probability theory. Since emergency calls arrive at rate λ = 0.8 calls per hour, the time between consecutive calls follows an exponential distribution with the same parameter λ = 0.8. For an exponential random variable, the probability that the waiting time T exceeds some value t is given by P(T > t) = e^(-λt). Here, you want P(T > 2) = e^(-0.8 × 2) = e^(-1.6) ≈ 0.202, which matches answer choice B. Let's examine why the other options are incorrect. Choice A gives 1 - e^(-1.6) ≈ 0.798, which represents P(T ≤ 2) — the probability that the time between calls is at most 2 hours, not exceeding 2 hours. This is the complement of what we want. Choice C shows e^(-0.8) ≈ 0.449, which would be correct if we were asking about waiting time exceeding 1 hour instead of 2 hours. Choice D gives 0.8 × 2 × e^(-1.6) ≈ 0.323, which incorrectly multiplies by the rate and time, perhaps confusing this with a Poisson probability mass function. Remember this key relationship: in a Poisson process with rate λ, inter-arrival times are exponentially distributed with the same rate parameter. Always check whether the question asks for "at most" versus "exceeds" — they're complements of each other.