Binomial Distribution Exercises in R: 10 Practice Problems Solved

These 10 binomial distribution exercises in R walk you from dbinom() one-liners to binom.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 binomial problems?

The binomial distribution answers one question: given n trials with success probability p, how likely is each count of successes? R gives you four functions — dbinom(), pbinom(), qbinom(), rbinom() — 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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# Fair coin, 10 flips — two classic questions dbinom(5, size = 10, prob = 0.5) # P(exactly 5 heads) #> [1] 0.2460938 pbinom(5, size = 10, prob = 0.5) # P(at most 5 heads) #> [1] 0.6230469

The first call asks "what is the probability of landing on exactly 5 heads?" and the second asks "what is the probability of 5 or fewer?" Both share the same parameters — size = 10 trials, prob = 0.5 success rate — and differ only in the function name. That two-function pairing handles most questions you will meet.

Decision tree: the four binomial functions.

Figure 1: The four binomial 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, ppois, qexp, rt). Learn it once for binomial, reuse it forever.

Try it: Compute the probability of exactly 3 successes in 8 trials with success probability 0.4.

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# Try it: exact probability with dbinom ex_p <- dbinom(___, size = ___, prob = ___) ex_p #> Expected: about 0.2787

Click to reveal solution

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ex_p <- dbinom(3, size = 8, prob = 0.4) ex_p #> [1] 0.2786918

Explanation: dbinom(k, n, p) returns P(X = k). Here X is the number of successes in 8 independent trials with p = 0.4, and k = 3.

How do you compute exact binomial probabilities with dbinom()?

dbinom(k, size = n, prob = p) returns the probability mass at a single count. The formula under the hood is the classic binomial PMF:

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$

Where:

$n$ = number of trials
$k$ = count of successes you ask about
$p$ = probability of success on each trial
$\binom{n}{k}$ = the number of ways to choose which trials succeed

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

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# Fair coin, 10 flips p_seven <- dbinom(7, size = 10, prob = 0.5) # exactly 7 heads p_seven #> [1] 0.1171875 # Between 3 and 5 heads, inclusive p_range <- sum(dbinom(3:5, size = 10, prob = 0.5)) p_range #> [1] 0.568359375

p_seven tells you that landing on exactly 7 heads happens about 12% of the time — not rare, but not the modal outcome either. The range call sums the PMF at k = 3, 4, 5. Passing the vector 3:5 vectorizes automatically, and sum() collapses the three probabilities into one number (roughly 0.57).

Tip

Vectorize dbinom for ranges instead of calling it in a loop. sum(dbinom(3:5, 10, 0.5)) is faster and more readable than three separate calls, and it scales cleanly when the range gets wider (e.g., sum(dbinom(30:60, 100, 0.5))).

Try it: Find the probability of exactly 4 successes in 12 trials with success probability 0.3.

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# Try it: dbinom with n=12, p=0.3 ex_p4 <- dbinom(___, size = ___, prob = ___) ex_p4 #> Expected: about 0.2311

Click to reveal solution

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ex_p4 <- dbinom(4, size = 12, prob = 0.3) ex_p4 #> [1] 0.2311397

Explanation: Same function, new parameters. The peak of this PMF sits near np = 3.6, so k = 4 is close to the mode and picks up a large share of the mass.

How do you compute cumulative probabilities with pbinom()?

pbinom(q, size = n, prob = p) returns $P(X \le q)$ — the cumulative probability up to and including q. "At least" questions flip the direction, and there are two equivalent ways to compute them: the explicit complement 1 - pbinom(...) or the lower.tail = FALSE argument.

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# n = 10, p = 0.3 p_leq3 <- pbinom(3, size = 10, prob = 0.3) # P(X <= 3) p_leq3 #> [1] 0.6496107 # P(X >= 7), two equivalent ways p_geq7a <- 1 - pbinom(6, size = 10, prob = 0.3) p_geq7a #> [1] 0.01059208 p_geq7b <- pbinom(6, size = 10, prob = 0.3, lower.tail = FALSE) p_geq7b #> [1] 0.01059208

p_leq3 says that in 10 trials with p = 0.3, three or fewer successes happen about 65% of the time. The two "at least 7" calls return the same ~1% probability: 1 - pbinom(6, ...) subtracts everything up to 6 from 1, while lower.tail = FALSE does the same arithmetic internally. Use whichever reads better in your code; the lower.tail form avoids a potential rounding issue when the tail is extremely small.

Warning

pbinom(q) is P(X ≤ q), not P(X < q). A common off-by-one mistake is to write pbinom(7, ...) when you really want "strictly less than 7" — the correct call is pbinom(6, ...). Always ask yourself whether the boundary is included before picking q.

Try it: For n = 5 and p = 0.2, find the probability of at least 2 successes.

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# Try it: P(X >= 2) with n=5, p=0.2 ex_ge <- 1 - pbinom(___, size = ___, prob = ___) ex_ge #> Expected: about 0.2627

Click to reveal solution

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ex_ge <- 1 - pbinom(1, size = 5, prob = 0.2) ex_ge #> [1] 0.26272 # Equivalent with lower.tail pbinom(1, size = 5, prob = 0.2, lower.tail = FALSE) #> [1] 0.26272

Explanation: "At least 2" means X ≥ 2, which is the complement of X ≤ 1. Subtract pbinom(1, ...) from 1, or use lower.tail = FALSE to get the upper tail directly.

How do you find quantiles and simulate with qbinom() and rbinom()?

qbinom() inverts pbinom(): given a cumulative probability, it returns the smallest count whose cumulative probability reaches that threshold. rbinom() draws random binomial counts — useful for simulation, Monte Carlo checks, and bootstrapping. The first argument of rbinom() is the number of random draws to make, not the trial count; the trial count is size.

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# n = 100 trials, p = 0.4 — 95th percentile of the success count q95 <- qbinom(0.95, size = 100, prob = 0.4) q95 #> [1] 48 # Five simulated outcomes from the same distribution set.seed(101) sim5 <- rbinom(5, size = 100, prob = 0.4) sim5 #> [1] 38 38 42 39 42

q95 = 48 tells you that in 95% of experiments you should see 48 or fewer successes, so 48 is a reasonable "worst-case-except-for-5%" threshold. The rbinom() call generates 5 independent samples from Binomial(100, 0.4); each number is one simulated experiment's success count, and they cluster around the mean $np = 40$. set.seed() ensures your results match when you rerun the code.

Note

rbinom's first argument is the number of draws, not the trial count. rbinom(5, 100, 0.4) means "give me 5 independent counts, each from Binomial(100, 0.4)" — not "5 trials with 100 draws each." Swapping the two is the most common bug with this function.

Try it: Simulate 10 draws from Binomial(20, 0.6) with set.seed(7) and compute the mean of the draws.

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# Try it: 10 simulated counts, their mean set.seed(7) ex_sim <- rbinom(___, size = ___, prob = ___) mean(ex_sim) #> Expected: a value near 12 (theoretical mean is np = 12)

Click to reveal solution

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set.seed(7) ex_sim <- rbinom(10, size = 20, prob = 0.6) ex_sim #> [1] 13 13 10 13 14 11 13 13 13 14 mean(ex_sim) #> [1] 12.7

Explanation: With only 10 draws, the sample mean (12.7) wanders a bit around the theoretical mean np = 12. Increase the number of draws to tighten the estimate — try 10,000 and compare.

Practice Exercises

Ten problems, progressively harder. Each one gives the setup, a starter block with a hint, and a hidden solution you can reveal after you try. Variables are prefixed ans or named by exercise so they do not clash with tutorial state.

Exercise 1: Exactly k heads in a fair coin toss

A fair coin is tossed 6 times. What is the probability of exactly 4 heads? Save the answer to ans1.

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# Exercise 1: exact probability with dbinom # Hint: size = 6, prob = 0.5 # Write your code below:

Click to reveal solution

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ans1 <- dbinom(4, size = 6, prob = 0.5) ans1 #> [1] 0.234375

Explanation: P(X = 4) for Binomial(6, 0.5). Four heads in six tosses is close to the mode (3), so the probability is a healthy ~23%.

Exercise 2: At most 2 defects in a batch

A factory produces parts with a 15% defect rate. In a batch of 20 parts, what is the probability of at most 2 defects? Save to ans2.

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# Exercise 2: cumulative probability with pbinom # Hint: "at most 2" means X <= 2 # Write your code below:

Click to reveal solution

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ans2 <- pbinom(2, size = 20, prob = 0.15) ans2 #> [1] 0.4048607

Explanation: P(X ≤ 2) in Binomial(20, 0.15). About 40% of batches pass a "2 defects or fewer" gate.

Exercise 3: At least 3 correct by guessing

A quiz has 10 multiple-choice questions, each with 4 options. What is the probability of getting at least 3 correct by guessing randomly? Save to ans3.

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# Exercise 3: complement with pbinom # Hint: p = 1/4, "at least 3" is the complement of "at most 2" # Write your code below:

Click to reveal solution

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ans3 <- 1 - pbinom(2, size = 10, prob = 0.25) ans3 #> [1] 0.4744072 # Equivalent with lower.tail pbinom(2, size = 10, prob = 0.25, lower.tail = FALSE) #> [1] 0.4744072

Explanation: With p = 0.25 and 10 questions, random guessing beats the "3 correct" bar almost half the time. The expected score is np = 2.5, which sits right on the boundary.

Exercise 4: Between 6 and 9 "yes" responses

In a survey of 12 people, each person says "yes" with probability 0.6. What is the probability of between 6 and 9 "yes" responses, inclusive? Save to ans4.

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# Exercise 4: sum over a range of k values # Hint: sum(dbinom(6:9, ...)) # Write your code below:

Click to reveal solution

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ans4 <- sum(dbinom(6:9, size = 12, prob = 0.6)) ans4 #> [1] 0.6988104

Explanation: Passing 6:9 to dbinom() returns a length-4 vector of PMF values; sum() adds them. The bulk of the mass in Binomial(12, 0.6) lives in this range because np = 7.2.

Exercise 5: Inventory stocking to cover 99% of demand

A store sees 200 customers per day, and each buys a niche product with probability 0.05. What is the smallest stock level that covers at least 99% of demand? Save to ans5.

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# Exercise 5: qbinom — smallest k with P(X <= k) >= 0.99 # Hint: qbinom(0.99, size = 200, prob = 0.05) # Write your code below:

Click to reveal solution

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ans5 <- qbinom(0.99, size = 200, prob = 0.05) ans5 #> [1] 18 # Verify pbinom(18, size = 200, prob = 0.05) #> [1] 0.9908445 pbinom(17, size = 200, prob = 0.05) #> [1] 0.9838987

Explanation: qbinom(0.99, ...) returns 18 — the smallest stock level whose cumulative probability first reaches 99%. Stocking 17 would only cover ~98.4%, so qbinom() gives you the right safety margin with a single call.

Exercise 6: Simulate dice rolls and compare with theory

Simulate 10,000 experiments of rolling a fair die 50 times, counting rolls that show a 6 (so p = 1/6). Use set.seed(42). Report the simulated mean and standard deviation, and compare with theoretical values $np$ and $\sqrt{np(1-p)}$. Save the vector to sim6.

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# Exercise 6: rbinom simulation at scale # Hint: rbinom(10000, size = 50, prob = 1/6) # Write your code below:

Click to reveal solution

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set.seed(42) sim6 <- rbinom(10000, size = 50, prob = 1/6) # Simulated summary c(mean = mean(sim6), sd = sd(sim6)) #> mean sd #> 8.320400 2.637789 # Theoretical n <- 50; p <- 1/6 c(np = n * p, sqrt_npq = sqrt(n * p * (1 - p))) #> np sqrt_npq #> 8.333333 2.635231

Explanation: Ten thousand draws get the sample mean within 0.02 of $np$ and the sample SD within 0.003 of $\sqrt{np(1-p)}$. This is why Monte Carlo checks are a reliable sanity test for any binomial calculation — the simulation converges fast.

Exercise 7: Mean and variance of a clinical trial response count

In a clinical trial, n = 30 patients each respond to treatment with probability 0.7. Compute the theoretical mean $E[X] = np$ and variance $Var[X] = np(1-p)$. Then draw 10,000 samples with set.seed(11) and compare.

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# Exercise 7: theory vs simulation for mean and variance # Hint: mean(rbinom(10000, 30, 0.7)) should be near np # Write your code below:

Click to reveal solution

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n <- 30; p <- 0.7 # Theory c(mean = n * p, variance = n * p * (1 - p)) #> mean variance #> 21.0 6.3 # Simulation set.seed(11) sim7 <- rbinom(10000, size = n, prob = p) c(sim_mean = mean(sim7), sim_var = var(sim7)) #> sim_mean sim_var #> 21.02120 6.36147

Explanation: Theory gives mean = 21 and variance = 6.3. The 10k simulated sample reproduces both to two decimal places. For any Binomial(n, p), the mean is always $np$ and the variance is always $np(1-p)$ — these two formulas are worth memorizing.

Exercise 8: Plot the PMF with a barplot

Plot the probability mass function of Binomial(n = 25, p = 0.4) over the counts 0 through 25. Use barplot(). Save the PMF vector to pmf8.

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# Exercise 8: barplot of dbinom over 0:25 # Hint: pmf8 <- dbinom(0:25, 25, 0.4) # Write your code below:

Click to reveal solution

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pmf8 <- dbinom(0:25, size = 25, prob = 0.4) barplot(pmf8, names.arg = 0:25, xlab = "Number of successes (k)", ylab = "P(X = k)", main = "PMF of Binomial(25, 0.4)", col = "steelblue") # The mode which.max(pmf8) - 1 #> [1] 10

Explanation: The mode sits at k = 10 (close to np = 10). barplot() accepts the PMF vector directly; names.arg maps each bar to its k value. A quick which.max() - 1 converts the 1-indexed vector position to the 0-indexed count.

Exercise 9: A/B test with binom.test

An A/B test logs 58 conversions in 100 visits. Test the null hypothesis that the true conversion rate is 0.5 against the two-sided alternative, and report the p-value and 95% confidence interval. Save the test object to test9.

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# Exercise 9: exact binomial test # Hint: binom.test(58, 100, p = 0.5) # Write your code below:

Click to reveal solution

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test9 <- binom.test(58, 100, p = 0.5) test9 #> #> Exact binomial test #> #> data: 58 and 100 #> number of successes = 58, number of trials = 100, p-value = 0.1332 #> alternative hypothesis: true probability of success is not equal to 0.5 #> 95 percent confidence interval: #> 0.4771192 0.6780145 #> sample estimates: #> probability of success #> 0.58

Explanation: The p-value of 0.133 sits above the usual 0.05 threshold, so 58/100 is consistent with a true rate of 0.5. The 95% CI of [0.477, 0.678] also contains 0.5 — both signals agree that the observed lift is not statistically significant at this sample size.

Exercise 10: Airline overbooking risk

An airline sells 110 tickets for a plane with 100 seats. Each passenger shows up independently with probability 0.9. What is the probability more than 100 passengers show up (forcing someone off)? Then compute the same probability if the airline sells 115 tickets instead. Save both to ans10a and ans10b.

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# Exercise 10: applied overbooking # Hint: P(X > 100) = 1 - pbinom(100, size, 0.9) # Write your code below:

Click to reveal solution

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ans10a <- 1 - pbinom(100, size = 110, prob = 0.9) ans10a #> [1] 0.3469894 ans10b <- 1 - pbinom(100, size = 115, prob = 0.9) ans10b #> [1] 0.8043866

Explanation: At 110 tickets, the airline faces a ~35% chance of having to bump passengers — already uncomfortable. Selling 115 tickets pushes that risk to ~80%, which almost guarantees a daily overbooking incident. Use qbinom() to pick a safer ticket count for a target risk level.

Complete Example: End-to-End Quality Control Workflow

A factory inspects batches of 500 widgets. Historically, 2% of widgets are defective. Walk through the four binomial functions on one coherent dataset: expected defects, tail probability, 95th percentile, simulation, and hypothesis test.

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# Setup batch_n <- 500 p_def <- 0.02 # 1. Expected defects (mean) and SD c(mean = batch_n * p_def, sd = sqrt(batch_n * p_def * (1 - p_def))) #> mean sd #> 10.00 3.13 # 2. Tail probability: P(defects > 15) 1 - pbinom(15, size = batch_n, prob = p_def) #> [1] 0.04908846 # 3. 95th percentile of defect count q95_def <- qbinom(0.95, size = batch_n, prob = p_def) q95_def #> [1] 15 # 4. Simulate 10,000 batches set.seed(2026) qc_sim <- rbinom(10000, size = batch_n, prob = p_def) mean(qc_sim > 15) # empirical tail probability #> [1] 0.0494 summary(qc_sim) #> Min. 1st Qu. Median Mean 3rd Qu. Max. #> 1.00 8.00 10.00 10.02 12.00 25.00 # 5. Hypothesis test: an inspector found 14 defects in 500 — # is this evidence the defect rate exceeds 2%? qc_test <- binom.test(14, 500, p = 0.02, alternative = "greater") qc_test$p.value #> [1] 0.1390315

The batch averages 10 defects with SD ~3.1, so 14 defects is one-plus standard deviation above the mean — noticeable but not alarming. pbinom() and the simulated tail both give about 5% for P(defects > 15); the simulation (4.94%) matches the analytic value (4.91%) within Monte Carlo error. qbinom() pins the 95th percentile at 15 defects. The one-sided binom.test() returns p = 0.139, confirming that 14/500 is not strong evidence of an elevated defect rate. This is the full analyst loop — point estimate, tail risk, quantile, simulation, formal test — in a dozen lines of R.

Summary

Question asked	Function	Syntax
P(X = k) exact	`dbinom()`	`dbinom(k, size = n, prob = p)`
P(X ≤ q) cumulative	`pbinom()`	`pbinom(q, size = n, prob = p)`
P(X ≥ q) upper tail	`pbinom()` with `lower.tail = FALSE`	`pbinom(q-1, n, p, lower.tail = FALSE)`
Smallest k for P(X ≤ k) ≥ α	`qbinom()`	`qbinom(alpha, size = n, prob = p)`
Random counts	`rbinom()`	`rbinom(N_draws, size = n, prob = p)`
Range sum	vectorize dbinom	`sum(dbinom(a:b, n, p))`
Hypothesis test	`binom.test()`	`binom.test(x, n, p = p0)`

The mean of Binomial(n, p) is always $np$ and the variance is always $np(1-p)$. The four functions share one prefix pattern (d/p/q/r) with every other distribution in base R, so the muscle memory you build here transfers to dnorm, ppois, qexp, and beyond.

References

R Core Team — ?Binomial reference manual. Link
Wickham, H. & Grolemund, G. — R for Data Science, 2nd ed. Link
Dalgaard, P. — Introductory Statistics with R, 2nd ed. Springer (2008).
Donovan, T., Coggins, L. & Hines, J. — Binomial Distribution in R. UVM (2020). Link
CRAN Task View: Distributions. Link
Rice, J. A. — Mathematical Statistics and Data Analysis, 3rd ed. Duxbury (2006).
Diez, D., Çetinkaya-Rundel, M. & Barr, C. — OpenIntro Statistics, 4th ed. Link
Butler, G. — Tutorial 4: The Binomial Distribution, ECON 41 Labs. Link

Continue Learning

Binomial vs Poisson in R: Understand When Each Distribution Fits Your Counts — the core tutorial these exercises extend.
Normal Distribution in R — the continuous cousin, also the large-n approximation for the binomial.
Probability Distributions in R — the bigger picture of the d/p/q/r family across distributions.

Binomial Distribution Exercises in R: 10 Practice Problems Solved

What quick R one-liners solve binomial problems?

How do you compute exact binomial probabilities with dbinom()?

How do you compute cumulative probabilities with pbinom()?

How do you find quantiles and simulate with qbinom() and rbinom()?

Practice Exercises

Exercise 1: Exactly k heads in a fair coin toss

Exercise 2: At most 2 defects in a batch

Exercise 3: At least 3 correct by guessing

Exercise 4: Between 6 and 9 "yes" responses

Exercise 5: Inventory stocking to cover 99% of demand

Exercise 6: Simulate dice rolls and compare with theory

Exercise 7: Mean and variance of a clinical trial response count

Exercise 8: Plot the PMF with a barplot

Exercise 9: A/B test with binom.test

Exercise 10: Airline overbooking risk

Complete Example: End-to-End Quality Control Workflow

Summary

References

Continue Learning

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