Poisson Distribution Exercises in R: 10 Problems with Solutions

These 10 Poisson distribution exercises in R walk you from dpois() one-liners to poisson.test() inference, with full runnable solutions. Every problem has a scaffold, a hint, and a reveal so you can check your answer the moment you finish coding.

What quick R one-liners solve Poisson problems?

The Poisson distribution answers one question: given an average rate λ of events in a window, how likely is each count? R gives you four functions — dpois(), ppois(), qpois(), rpois() — that cover exact probability, cumulative probability, quantiles, and random samples. Before the 10 problems, a single runnable block shows the two most common cases side by side so you can spot which function fits a new question at a glance.

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# Call centre — 4 calls/hour on average dpois(3, lambda = 4) # P(exactly 3 calls in 1 hour) #> [1] 0.1953668 ppois(3, lambda = 4) # P(at most 3 calls in 1 hour) #> [1] 0.4334701

The first call asks "what is the probability of exactly 3 calls?" and the second asks "what is the probability of 3 or fewer?" Both share the same parameter — lambda = 4 events per hour — and differ only in the function name. That two-function pairing handles most questions you will meet.

Decision tree: the four Poisson functions.

Figure 1: The four Poisson functions — pick one by what the question asks for.

Key Insight

R's d/p/q/r prefix pattern is universal. The same four-letter scheme — d for density/mass, p for cumulative, q for quantile, r for random — works for every distribution in base R (dnorm, pbinom, qexp, rt). Learn it once for Poisson, reuse it forever.

Try it: Compute the probability of exactly 2 events when the average rate is λ = 5.

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# Try it: exact probability with dpois ex_p <- dpois(___, lambda = ___) ex_p #> Expected: about 0.0842

Click to reveal solution

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ex_p <- dpois(2, lambda = 5) ex_p #> [1] 0.08422434

Explanation: dpois(k, lambda) returns P(X = k). With λ = 5 events on average but k = 2 observed, the count falls well below the mean, so the probability is small.

How do you compute exact Poisson probabilities with dpois()?

dpois(k, lambda) returns the probability mass at a single count. The formula under the hood is the classic Poisson PMF:

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

Where:

$\lambda$ = the average rate of events per window
$k$ = the count of events you ask about
$e$ = Euler's constant, ≈ 2.71828
$k!$ = k factorial (the number of orderings of k items)

For a single k, one dpois() call is enough. For a range of counts — say "between 3 and 5 emails" — pass a vector of k values and sum the result.

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# Emails — 3 per hour on average dpois(5, lambda = 3) # P(exactly 5 emails) #> [1] 0.1008188 p_emails <- sum(dpois(3:5, lambda = 3)) # P(3, 4, or 5 emails) p_emails #> [1] 0.4901484

The first call gives a direct mass at k = 5. The second passes a vector 3:5 to dpois(), returning three probabilities in one go, which sum() collapses into the total for the range. The answer — about 49% — is the probability that the hour lands anywhere in that 3-to-5-email band.

Tip

Pass a vector of k values to vectorize dpois(). For any "between a and b" range question, sum(dpois(a:b, lambda)) is the one-line idiom. It is faster and clearer than a loop, and it beats typing three separate dpois() calls.

Try it: A bus arrives at a stop at a rate of 1.5 per 10 minutes. Compute P(exactly 2 arrivals in a 10-minute window).

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# Try it: bus arrivals with dpois ex_bus <- dpois(___, lambda = ___) ex_bus #> Expected: about 0.2510

Click to reveal solution

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ex_bus <- dpois(2, lambda = 1.5) ex_bus #> [1] 0.2510214

Explanation: With λ = 1.5 buses per 10 minutes, seeing exactly 2 is close to the mean, so the probability is high — about 25%.

How do you compute cumulative Poisson probabilities with ppois()?

ppois(q, lambda) returns the cumulative probability up to and including q — that is, P(X ≤ q). For "at least" questions, use the complement rule or the lower.tail = FALSE shortcut.

Two ways to get "at least 7":

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# Customer arrivals — 5 per hour on average p_at_most <- ppois(5, lambda = 5) p_at_most #> [1] 0.6159607 p_at_least <- 1 - ppois(6, lambda = 5) # complement p_at_least #> [1] 0.2378398 ppois(6, lambda = 5, lower.tail = FALSE) # same thing, cleaner #> [1] 0.2378398

The first line returns the probability of 5 or fewer arrivals — just over 61%, a bit more than half since 5 matches the mean exactly. The next two lines compute P(X ≥ 7) by subtracting P(X ≤ 6) from 1. The lower.tail = FALSE form skips the subtraction and returns the upper tail directly — the idiom to reach for in every "at least" problem.

Warning

ppois(q) returns P(X ≤ q), NOT P(X < q). This is the most common off-by-one trap in Poisson problems. For strict inequality P(X < 7), pass q = 6 to ppois(). For P(X ≥ 7), pass q = 6 with lower.tail = FALSE — not q = 7.

Try it: Defects arrive on a production line at λ = 2 per shift. Compute P(at least 3 defects in a shift).

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# Try it: at-least probability with ppois ex_defects <- ppois(___, lambda = ___, lower.tail = ___) ex_defects #> Expected: about 0.3233

Click to reveal solution

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ex_defects <- ppois(2, lambda = 2, lower.tail = FALSE) ex_defects #> [1] 0.3233236

Explanation: P(X ≥ 3) is the upper tail starting strictly above 2. Pass q = 2 with lower.tail = FALSE — ppois() returns P(X > 2), which is the same as P(X ≥ 3) for an integer-valued distribution.

How do you find quantiles and simulate counts with qpois() and rpois()?

qpois(p, lambda) inverts ppois() — give it a cumulative probability and it returns the smallest count k where P(X ≤ k) reaches p. rpois(n, lambda) draws n random counts from the distribution, useful for simulation.

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# Website traffic — 10 visits/minute on average q95 <- qpois(0.95, lambda = 10) # 95th-percentile traffic q95 #> [1] 15 set.seed(42) sim_counts <- rpois(5, lambda = 10) # five simulated minutes sim_counts #> [1] 9 14 10 13 9

The first call answers "what is the smallest visit count that covers 95% of minutes?" — 15 visits. If you provisioned a server for 15 concurrent visits, it would handle 95% of one-minute windows without overload. The second call draws 5 random counts from the same distribution, which hover around the mean of 10 — exactly what a real minute-by-minute log would look like.

Note

rpois's first argument is the number of draws, not lambda. Write rpois(n = 1000, lambda = 5) for 1,000 random counts, not rpois(5, 1000). The order matters — unnamed arguments swap roles silently.

Try it: Set set.seed(7) and simulate 10 counts with λ = 4. Report the sample mean.

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# Try it: simulate and report mean set.seed(7) ex_sim <- rpois(___, lambda = ___) mean(ex_sim) #> Expected: around 4 (sampling noise varies each run)

Click to reveal solution

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set.seed(7) ex_sim <- rpois(10, lambda = 4) ex_sim #> [1] 6 3 3 4 8 5 3 6 3 4 mean(ex_sim) #> [1] 4.5

Explanation: The theoretical mean of a Poisson(λ) is λ itself. A 10-draw sample will land close but not exactly on 4 because of sampling noise; a 10,000-draw sample would converge much tighter.

Practice Exercises

Ten problems, arranged from easy to hard. Each uses distinct my_* variable names so your exercise work does not overwrite the tutorial state above. Write your solution in the starter block, run it, then open the reveal to check.

Exercise 1: Emergency room admissions (exact count)

A hospital emergency room averages 6 admissions per hour. Compute P(exactly 4 admissions in an hour). Save the answer to my_p1.

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# Exercise 1: exact count with dpois # Hint: dpois(k, lambda) gives P(X = k) my_p1 <- print(my_p1) #> Expected: about 0.1339

Click to reveal solution

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my_p1 <- dpois(4, lambda = 6) print(my_p1) #> [1] 0.1338526

Explanation: With λ = 6 per hour but k = 4, the observed count is below the mean, so the probability dips below the mass at k = 6 (which is the mode).

Exercise 2: Monthly website outages (at most k)

A website averages 2 outages per month. Compute P(at most 1 outage in a given month). Save to my_p2.

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# Exercise 2: cumulative probability with ppois # Hint: "at most k" means P(X <= k) my_p2 <- print(my_p2) #> Expected: about 0.4060

Click to reveal solution

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my_p2 <- ppois(1, lambda = 2) print(my_p2) #> [1] 0.4060058

Explanation: ppois(1, 2) sums the masses at 0 and 1 outages — about 41% of months stay within the "0 or 1" window.

Exercise 3: Product defects (at least 2)

A production unit averages 0.8 defects per unit. Compute P(at least 2 defects in a unit). Save to my_p3.

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# Exercise 3: complement with ppois # Hint: use lower.tail = FALSE and pass q = 1 my_p3 <- print(my_p3) #> Expected: about 0.1912

Click to reveal solution

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my_p3 <- ppois(1, lambda = 0.8, lower.tail = FALSE) print(my_p3) #> [1] 0.1912079

Explanation: P(X ≥ 2) = P(X > 1). Pass q = 1 with lower.tail = FALSE — this is the upper-tail idiom that replaces the manual 1 - ppois(...) subtraction.

Exercise 4: Insurance claims (range)

An insurer receives 5 claims per day on average. Compute P(between 3 and 7 claims, inclusive). Save to my_p4.

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# Exercise 4: range using sum(dpois(a:b, lambda)) # Hint: 3:7 is a vector of 5 values my_p4 <- print(my_p4) #> Expected: about 0.7419

Click to reveal solution

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my_p4 <- sum(dpois(3:7, lambda = 5)) print(my_p4) #> [1] 0.7419466

Explanation: Passing 3:7 to dpois() returns five probabilities, one per count. sum() collapses them into the total mass for the inclusive range — about 74%, a large share because the range is centred on the mean.

Exercise 5: Helpdesk staffing (quantile)

A helpdesk receives 12 calls per hour on average. Find the smallest staff count my_staff such that P(calls ≤ my_staff) ≥ 0.95 — enough capacity to cover 95% of hours.

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# Exercise 5: inverse CDF with qpois # Hint: qpois returns the smallest k with P(X <= k) >= p my_staff <- print(my_staff) #> Expected: 18

Click to reveal solution

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my_staff <- qpois(0.95, lambda = 12) print(my_staff) #> [1] 18 # Sanity check ppois(18, lambda = 12) #> [1] 0.9625197

Explanation: qpois(0.95, 12) answers "what staff count covers at least 95% of hours?" — 18. The sanity check confirms ppois(18, 12) indeed exceeds 0.95, while ppois(17, 12) does not.

Exercise 6: Call-centre simulation (rpois + summary stats)

Simulate 10,000 hours of a call centre with λ = 8. Set set.seed(44). Save the draws to my_sim6 and compute the sample mean and variance — both should land near 8.

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# Exercise 6: simulate and verify mean = variance = lambda set.seed(44) my_sim6 <- c(sample_mean = mean(my_sim6), sample_var = var(my_sim6)) #> Expected: both about 8

Click to reveal solution

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set.seed(44) my_sim6 <- rpois(10000, lambda = 8) c(sample_mean = mean(my_sim6), sample_var = var(my_sim6)) #> sample_mean sample_var #> 7.985300 7.949594

Explanation: A Poisson distribution has the unusual property that E[X] = Var[X] = λ. A 10,000-draw sample makes both estimates snap tight to 8 — a quick visual check that your data really is Poisson-shaped.

Exercise 7: Bus arrivals (rate scaling)

Buses arrive at a stop at 2 per 15 minutes on average. Convert the rate to arrivals per hour (should be λ = 8) and compute P(at least 10 buses in a 60-minute window). Save to my_p7.

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# Exercise 7: scale lambda, then compute upper tail # Hint: 2 per 15 min = 8 per 60 min; then use ppois with lower.tail = FALSE my_lambda7 <- my_p7 <- print(my_p7) #> Expected: about 0.2834

Click to reveal solution

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my_lambda7 <- 2 * 4 # 4 quarters in an hour => lambda = 8 / hour my_p7 <- ppois(9, lambda = my_lambda7, lower.tail = FALSE) print(my_p7) #> [1] 0.2833757

Explanation: Poisson rates scale linearly with the time window. Doubling the window doubles λ. Once rescaled to hourly units, this becomes a standard "at least 10" upper-tail query, so use lower.tail = FALSE with q = 9.

Exercise 8: Plot the PMF (visualization)

Plot the probability mass function of Poisson(λ = 7) over the range 0:20 using barplot(). Label the axes and highlight the mode bar in a distinct colour.

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# Exercise 8: PMF barplot # Hint: dpois(0:20, 7) gives you the full vector of masses my_x8 <- 0:20 my_masses8 <- my_colours8 <- barplot(my_masses8, names.arg = my_x8, col = my_colours8, xlab = "k (count)", ylab = "P(X = k)", main = "Poisson PMF, lambda = 7")

Click to reveal solution

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my_x8 <- 0:20 my_masses8 <- dpois(my_x8, lambda = 7) my_colours8 <- ifelse(my_masses8 == max(my_masses8), "steelblue", "grey70") barplot(my_masses8, names.arg = my_x8, col = my_colours8, xlab = "k (count)", ylab = "P(X = k)", main = "Poisson PMF, lambda = 7")

Explanation: The Poisson PMF peaks at the integer floor of λ — here, k = 7. The highlighted bar makes that mode obvious. For non-integer λ, two adjacent bars can share the maximum.

Exercise 9: Rate test (poisson.test)

Over 10 time units you observed 25 events. Test the null hypothesis H0: λ = 2 against a two-sided alternative using poisson.test(). Save the result to my_test9 and extract the p-value.

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# Exercise 9: exact rate test # Hint: poisson.test(x, T, r) where x = observed, T = time, r = null rate my_test9 <- my_test9$p.value #> Expected: about 0.0343

Click to reveal solution

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my_test9 <- poisson.test(x = 25, T = 10, r = 2) print(my_test9) #> Exact Poisson test #> #> data: 25 time base: 10 #> number of events = 25, time base = 10, p-value = 0.03426 #> alternative hypothesis: true event rate is not equal to 2 #> 95 percent confidence interval: #> 1.617477 3.691057 #> sample estimate: #> event rate #> 2.5 my_test9$p.value #> [1] 0.03425515

Explanation: The observed rate is 2.5 events per unit, higher than the hypothesised 2. The exact two-sided p-value is about 0.034, so you reject H0 at α = 0.05. The 95% CI (1.62, 3.69) excludes 2 at the lower bound only just barely — the evidence is there, but not overwhelming.

Exercise 10: Hospital bed planning (capacity + simulation)

A hospital admits 30 patients per day on average. What is the smallest bed capacity my_beds such that the probability of overflow — P(admissions > my_beds) — stays below 5%? Verify with a 10,000-day simulation (set.seed(2026)): the empirical overflow share should sit under 0.05.

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# Exercise 10: capacity planning, then verify by simulation # Hint: use qpois(0.95, 30), then check with rpois my_beds <- print(my_beds) #> Expected: 39 set.seed(2026) my_days10 <- rpois(10000, lambda = 30) my_overflow <- mean(my_days10 > my_beds) print(my_overflow) #> Expected: around 0.03

Click to reveal solution

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my_beds <- qpois(0.95, lambda = 30) print(my_beds) #> [1] 39 set.seed(2026) my_days10 <- rpois(10000, lambda = 30) my_overflow <- mean(my_days10 > my_beds) print(my_overflow) #> [1] 0.0299

Explanation: qpois(0.95, 30) returns 39 — the smallest count with cumulative probability ≥ 95%. The simulation confirms only about 3% of days exceed 39 admissions, well within the 5% overflow budget. If you tried 38 beds instead, the empirical overflow would jump above 5%.

Complete Example: End-to-End ER Admissions Workflow

A real capacity study combines every function you just practised. Walk through a full hour-by-hour ER admissions analysis: compute the theoretical moments, cap capacity at the 95th percentile, simulate a week of hours to sanity-check the distribution, then test the rate against a historical baseline using 72 hours of recent data.

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# Scenario: ER admissions, lambda = 6 per hour er_lambda <- 6 # 1. Theoretical mean and variance (both equal lambda for Poisson) c(mean = er_lambda, variance = er_lambda) #> mean variance #> 6 6 # 2. Probability the hour is a "stress hour" (8 or more admissions) ppois(7, lambda = er_lambda, lower.tail = FALSE) #> [1] 0.2560202 # 3. 95th-percentile hourly capacity target qpois(0.95, lambda = er_lambda) #> [1] 10 # 4. Simulate a week (168 hours) and summarise set.seed(2026) er_sim <- rpois(168, lambda = er_lambda) summary(er_sim) #> Min. 1st Qu. Median Mean 3rd Qu. Max. #> 0.000 4.000 6.000 5.976 8.000 12.000 # 5. Rate test: observed 520 admissions over 72 hours vs baseline lambda = 6 er_test <- poisson.test(x = 520, T = 72, r = 6) er_test$p.value #> [1] 0.0008842947 er_test$conf.int #> [1] 6.608921 7.866632 #> attr(,"conf.level") #> [1] 0.95

The five steps form a reproducible capacity review. First, the theoretical mean and variance both equal 6 — the Poisson signature. Second, about 26% of hours will cross 8 admissions, so roughly one in four hours is a stress hour. Third, qpois(0.95, 6) sets the 95th-percentile staffing target at 10 beds. Fourth, a 168-hour simulation returns a sample mean near 6 and a max of 12, exactly what a real week would look like. Fifth, the poisson.test() on 520 admissions in 72 hours returns a p-value of about 0.0009 and a 95% CI of (6.61, 7.87) — conclusive evidence that the recent rate has climbed above the baseline of 6, which should prompt a staffing review.

Summary

The table below is the full decision reference for Poisson problems in R. Pair it with the four-function picker in Figure 1 and the PMF formula to handle every exercise type you just practised.

Function	What it returns	Typical question	One-line idiom
`dpois(k, lambda)`	P(X = k)	"Exactly k events"	`dpois(5, 3)`
`ppois(q, lambda)`	P(X ≤ q)	"At most q events"	`ppois(5, 3)`
`ppois(q, lambda, lower.tail = FALSE)`	P(X > q)	"At least q+1 events"	`ppois(4, 3, lower.tail = FALSE)`
`qpois(p, lambda)`	smallest k with P(X ≤ k) ≥ p	"Capacity for p% of windows"	`qpois(0.95, 3)`
`rpois(n, lambda)`	n random counts	"Simulate n windows"	`rpois(1000, 3)`
`poisson.test(x, T, r)`	exact rate test + 95% CI	"Is the rate different from r?"	`poisson.test(25, 10, 2)`

Key facts to remember:

The PMF is $P(X = k) = \lambda^k e^{-\lambda} / k!$.
Mean and variance both equal λ — the Poisson fingerprint.
Poisson rates scale linearly with the window: doubling the window doubles λ.
ppois(q) returns P(X ≤ q), not P(X < q) — an off-by-one trap.
For ranges, sum(dpois(a:b, lambda)) is the vectorised idiom.

References

R Core Team — The Poisson Distribution (?Poisson help page). Link
Wickham, H. & Grolemund, G. — R for Data Science, 2nd Edition (2023). Link
Ross, S. — Introduction to Probability Models, 12th Edition. Academic Press (2019). Chapter 5: The Poisson Process.
Zach — A Guide to dpois, ppois, qpois, and rpois in R. Statology. Link
Soage, J. C. — Poisson Distribution in R. R-Coder. Link
Schork, J. — Poisson Distribution in R (4 Examples). Statistics Globe. Link
Rusczyk, D. — Exercises on the Poisson Distribution. Emory Math Center. Link
Dalgaard, P. — Introductory Statistics with R, 2nd Edition. Springer (2008). Chapter 3: Probability and Distributions.

Continue Learning

Binomial and Poisson Distributions in R — the core tutorial this exercise set sharpens. Read it first if dpois() vs dbinom() still feels blurry.
Binomial Distribution Exercises in R — sibling exercise set. Same scaffold-hint-reveal format, applied to dbinom/pbinom/qbinom/rbinom.
Random Variables in R — the foundational post covering PMFs, CDFs, and quantile functions that dpois/ppois/qpois implement.

Poisson Distribution Exercises in R: 10 Problems with Solutions

What quick R one-liners solve Poisson problems?

How do you compute exact Poisson probabilities with dpois()?

How do you compute cumulative Poisson probabilities with ppois()?

How do you find quantiles and simulate counts with qpois() and rpois()?

Practice Exercises

Exercise 1: Emergency room admissions (exact count)

Exercise 2: Monthly website outages (at most k)

Exercise 3: Product defects (at least 2)

Exercise 4: Insurance claims (range)

Exercise 5: Helpdesk staffing (quantile)

Exercise 6: Call-centre simulation (rpois + summary stats)

Exercise 7: Bus arrivals (rate scaling)

Exercise 8: Plot the PMF (visualization)

Exercise 9: Rate test (poisson.test)

Exercise 10: Hospital bed planning (capacity + simulation)

Complete Example: End-to-End ER Admissions Workflow

Summary

References

Continue Learning

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