Binomial Probability: Formula, Examples & Practice Problems

8 fully solved examples, 4 practice problems, and the mean/standard deviation

What is a Binomial Experiment?

A statistical experiment is a binomial experiment if it satisfies all four conditions:

  1. There is a fixed number of identical trials, denoted \( n \).
  2. Each trial has exactly two outcomes: success or failure.
  3. The probability of success, denoted \( p \), is the same on every trial.
  4. The trials are independent of one another.

Real-World Examples

Binomial Formulas

Quantity Formula
Probability of exactly \( k \) successes \( P(k) = \dbinom{n}{k}\, p^k\,(1-p)^{n-k} \)
Binomial coefficient (ways to choose \( k \) successes) \( \dbinom{n}{k} = \dfrac{n!}{k!\,(n-k)!} \)
Mean (expected number of successes) \( \mu = n \cdot p \)
Standard deviation \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \)

Intuition for the formula. The factor \( p^k (1-p)^{n-k} \) is the probability of one specific arrangement of \( k \) successes and \( n - k \) failures. The factor \( \binom{n}{k} \) counts how many distinct arrangements exist — because each valid arrangement contributes the same probability.

Factorial: \( n! = 1 \cdot 2 \cdot 3 \cdots n \), with \( 0! = 1 \).

Step-by-Step Worked Examples

Example 1 — 3 coin tosses, exactly 2 heads

A fair coin is tossed 3 times. What is the probability of getting exactly 2 heads?

View solution

The sample space for 3 tosses is:

Tree diagram of 3 coin tosses, showing all 8 possible sequences from HHH to TTT
Figure 1 — All 8 outcomes of tossing a fair coin 3 times.

The event "exactly 2 heads" is \( E = \{\text{HHT}, \text{HTH}, \text{THH}\} \), which has 3 outcomes.

Any specific sequence with 2 heads and 1 tail has probability:

\[ P(\text{HHT}) = p \cdot p \cdot (1-p) = p^2(1-p) \]

So for a fair coin (\( p = 0.5 \)):

\[ P(E) = 3 \cdot (0.5)^2 (0.5)^1 = 3 \cdot 0.125 = 0.375 \]

The coefficient 3 equals \( \binom{3}{2} \) — the number of ways to arrange 2 heads and 1 tail among 3 positions.

Example 2 — 5 tosses, exactly 3 heads

A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads?

View solution

Here \( n = 5 \), \( k = 3 \), \( p = 0.5 \).

\[ \binom{5}{3} = \frac{5!}{3!\,2!} = 10 \] \[ P(3) = 10 \cdot (0.5)^3 \cdot (0.5)^2 = 10 \cdot 0.125 \cdot 0.25 = 0.3125 \]

Example 3 — 7 die rolls, exactly 5 sixes

A fair die is rolled 7 times. Find the probability of rolling exactly 5 sixes.

View solution

Here \( n = 7 \), \( k = 5 \), \( p = 1/6 \).

\[ \binom{7}{5} = 21 \] \[ P(5) = 21 \left(\tfrac{1}{6}\right)^5 \left(\tfrac{5}{6}\right)^2 = 21 \cdot \frac{1}{7776} \cdot \frac{25}{36} \approx 0.00188 \]

Example 4 — Quality control mean and standard deviation

A factory produces tools with a 98% reliability rate. In a random sample of 1000 tools:

  1. Find the expected (mean) number of working tools.
  2. Find the standard deviation.
View solution

\( n = 1000 \), \( p = 0.98 \).

  1. Mean: \( \mu = n p = 1000 \cdot 0.98 = 980 \) tools.
  2. Standard deviation: \( \sigma = \sqrt{np(1-p)} = \sqrt{1000 \cdot 0.98 \cdot 0.02} = \sqrt{19.6} \approx 4.43 \).

Example 5 — 7 tosses, at least 5 heads

A fair coin is tossed 7 times. Find the probability of getting at least 5 heads.

View solution

"At least 5 heads" means \( P(5) + P(6) + P(7) \). Here \( n = 7 \), \( p = 0.5 \).

\[ P(5) = \binom{7}{5}(0.5)^7 = \frac{21}{128} = 0.16406 \] \[ P(6) = \binom{7}{6}(0.5)^7 = \frac{7}{128} = 0.05469 \] \[ P(7) = \binom{7}{7}(0.5)^7 = \frac{1}{128} = 0.00781 \] \[ P(\ge 5) = 0.16406 + 0.05469 + 0.00781 = 0.22656 \]

Example 6 — Passing a 20-question test by guessing

A test has 20 multiple-choice questions with 4 choices each. What is the probability of passing (scoring at least 10 correct) by random guessing?

View solution

\( n = 20 \), \( p = 0.25 \) (1 correct out of 4 choices).

\[ P(\ge 10) = \sum_{k=10}^{20} \binom{20}{k}(0.25)^k(0.75)^{20-k} \] \[ = 0.00992 + 0.00301 + 0.00075 + \cdots \approx 0.01386 \]

Answer: only about 1.4% — random guessing is highly ineffective.

Example 7 — 6 draws with replacement, at least 2 red

A box contains 3 red, 4 white, and 3 black balls. A ball is drawn 6 times with replacement. Find the probability of drawing at least 2 red balls.

View solution

\( p = \frac{3}{10} = 0.3 \), \( n = 6 \). Summing \( P(2) + \cdots + P(6) \) is tedious — use the complement rule:

\[ P(\ge 2) = 1 - \bigl[P(0) + P(1)\bigr] \] \[ P(0) = \binom{6}{0}(0.3)^0(0.7)^6 = 0.11765 \] \[ P(1) = \binom{6}{1}(0.3)^1(0.7)^5 = 0.30253 \] \[ P(\ge 2) = 1 - (0.11765 + 0.30253) = 1 - 0.42018 = 0.57982 \]

Example 8 — Insurance company sampling

80% of people in a city have their insurance with "MyInsurance".

  1. If 10 people are randomly selected, find the probability that at least 8 have insurance with this company.
  2. In a random sample of 500 people, how many are expected to have this insurance?
View solution

  1. \( n = 10 \), \( p = 0.8 \).

    \[ P(8) = \binom{10}{8}(0.8)^8(0.2)^2 \approx 0.30199 \] \[ P(9) = \binom{10}{9}(0.8)^9(0.2)^1 \approx 0.26844 \] \[ P(10) = \binom{10}{10}(0.8)^{10}(0.2)^0 \approx 0.10737 \] \[ P(\ge 8) = 0.30199 + 0.26844 + 0.10737 = 0.67780 \]
  2. Expected number: \( \mu = 500 \cdot 0.8 = 400 \) people.

Practice Problems

Practice 1 — 5 die rolls and even numbers

A fair die is rolled 5 times.

  1. Find the probability of rolling exactly 3 even numbers.
  2. Find the probability of rolling at least 3 even numbers.
  3. Find the probability of rolling at most 3 even numbers.
Answer

\( p = 3/6 = 0.5 \) (even numbers 2, 4, 6), \( n = 5 \).

  1. \( P(3) = \binom{5}{3}(0.5)^5 = 10/32 = 0.3125 \)
  2. \( P(\ge 3) = P(3) + P(4) + P(5) = 10/32 + 5/32 + 1/32 = 16/32 = 0.5 \)
  3. \( P(\le 3) = 1 - P(4) - P(5) = 1 - 5/32 - 1/32 = 26/32 = 0.8125 \)

Practice 2 — 10 draws with replacement, at least 3 red cards

A card is drawn from a standard deck (with replacement) 10 times. Find the probability of drawing at least 3 red cards.

Answer

\( p = 26/52 = 0.5 \), \( n = 10 \). Use the complement:

\[ P(\ge 3) = 1 - [P(0) + P(1) + P(2)] \] \[ = 1 - \bigl[(0.5)^{10} + 10(0.5)^{10} + 45(0.5)^{10}\bigr] \] \[ = 1 - \frac{56}{1024} = 1 - 0.05469 = 0.94531 \]

Practice 3 — Passing a 20-question test with 5 choices

A multiple-choice test has 20 questions with 5 choices each. What is the probability of passing (scoring at least 15 correct) solely by guessing?

Answer

\( n = 20 \), \( p = 0.20 \).

\[ P(\ge 15) = \sum_{k=15}^{20} \binom{20}{k}(0.2)^k(0.8)^{20-k} \]

The leading term is \( P(15) = \binom{20}{15}(0.2)^{15}(0.8)^5 \approx 1.66 \times 10^{-7} \), and every subsequent term is smaller still, so the entire sum is well under \( 10^{-6} \).

Conclusion: it is virtually impossible to pass this test by guessing.

Practice 4 — Expected counts from survey data

In the 25–34 age group, 61.8% of individuals in Canada and 50.8% in the UK have a tertiary education. If 200,000 people from this demographic are randomly selected from each country, how many are expected to have a tertiary education?

Answer

Use \( \mu = n p \):

  • Canada: \( 200{,}000 \times 0.618 = 123{,}600 \)
  • UK: \( 200{,}000 \times 0.508 = 101{,}600 \)

Frequently Asked Questions

What is a binomial experiment?

A binomial experiment has four properties: (1) a fixed number \(n\) of trials; (2) only two possible outcomes per trial (success or failure); (3) a constant probability of success \(p\) on each trial; (4) independent trials.

What is the binomial probability formula?

For exactly \(k\) successes in \(n\) independent trials with success probability \(p\): \(P(k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

What are the mean and standard deviation of a binomial distribution?

Mean: \(\mu = np\). Standard deviation: \(\sigma = \sqrt{np(1-p)}\). For example, with \(n = 1000\) and \(p = 0.98\), \(\mu = 980\) and \(\sigma = \sqrt{19.6} \approx 4.43\).

How do I compute an "at least" binomial probability efficiently?

For \(P(X \ge k)\) when \(k\) is greater than \(n/2\), use the complement rule: \(P(X \ge k) = 1 - P(X \le k-1)\). Summing fewer terms is faster and less error-prone. For \(P(X \le k)\) when \(k\) is small, sum directly. The same applies symmetrically.

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Last updated: October 20, 2026