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Bernoulli Trials — Definition, Formula & Examples

Bernoulli Trials

An experiment in which a single action, such as flipping a coin, is repeated identically over and over. The possible results of the action are classified as "success" or "failure". The binomial probability formula is used to find probabilities for Bernoulli trials.

Note: With Bernoulli trials, the repeated actions must all be independent.

 

Binomial Probability Formula:

P(k successes in n trials) = \(\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right){p^k}{q^{n - k}}\)

n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
Example:

You are taking a 10 question multiple choice test. If each question has four choices and you guess on each question, what is the probability of getting exactly 7 questions correct?

n = 10
k = 7
n – k = 3
p = 0.25 = probability of guessing the correct answer on a question
q = 0.75 = probability of guessing the wrong answer on a question

P(7 correct guesses in 10 questions) = \(\left( {\begin{array}{*{20}{c}}{10}\\7\end{array}} \right){\left( {0.25} \right)^7}{\left( {0.75} \right)^3} \approx 0.0031\)

See also

Binomial coefficients

Key Formula

P(k successes in n trials)=(nk)pkqn−kP(k \text{ successes in } n \text{ trials}) = \binom{n}{k} p^k q^{n-k}
Where:
  • nn = Total number of independent trials
  • kk = Number of successes you want
  • n−kn - k = Number of failures
  • pp = Probability of success on a single trial
  • qq = Probability of failure on a single trial, equal to 1 − p
  • (nk)\binom{n}{k} = Binomial coefficient — the number of ways to choose which k trials are successes

Worked Example

Problem: A fair coin is flipped 8 times. What is the probability of getting exactly 5 heads?
Step 1: Identify the parameters. Each flip is an independent trial with two outcomes (heads or tails), so these are Bernoulli Trials. Define "heads" as success.
n=8,k=5,p=0.5,q=1−0.5=0.5n = 8, \quad k = 5, \quad p = 0.5, \quad q = 1 - 0.5 = 0.5
Step 2: Calculate the binomial coefficient. This counts the number of ways to arrange 5 heads among 8 flips.
(85)=8!5!⋅3!=8×7×63×2×1=56\binom{8}{5} = \frac{8!}{5! \cdot 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
Step 3: Compute the probability of 5 successes and 3 failures in a specific order.
pk⋅qn−k=(0.5)5⋅(0.5)3=(0.5)8=1256p^k \cdot q^{n-k} = (0.5)^5 \cdot (0.5)^3 = (0.5)^8 = \frac{1}{256}
Step 4: Multiply the binomial coefficient by the single-order probability.
P(5 heads)=56×1256=56256=0.21875P(5 \text{ heads}) = 56 \times \frac{1}{256} = \frac{56}{256} = 0.21875
Answer: The probability of getting exactly 5 heads in 8 fair coin flips is approximately 0.219, or about 21.9%.

Another Example

This example differs from the first in two ways: the probability of success is not 0.5, and it asks for 'at least' a certain number of successes rather than 'exactly,' requiring you to sum multiple binomial terms.

Problem: A factory produces light bulbs with a 90% chance each bulb works (success). In a random sample of 4 bulbs, what is the probability that at least 3 bulbs work?
Step 1: Identify the parameters. Each bulb is an independent trial. Define "bulb works" as success.
n=4,p=0.9,q=0.1n = 4, \quad p = 0.9, \quad q = 0.1
Step 2: "At least 3" means k = 3 or k = 4. Calculate each probability separately and add them.
P(k≥3)=P(k=3)+P(k=4)P(k \geq 3) = P(k = 3) + P(k = 4)
Step 3: Find P(k = 3).
P(3)=(43)(0.9)3(0.1)1=4×0.729×0.1=0.2916P(3) = \binom{4}{3}(0.9)^3(0.1)^1 = 4 \times 0.729 \times 0.1 = 0.2916
Step 4: Find P(k = 4).
P(4)=(44)(0.9)4(0.1)0=1×0.6561×1=0.6561P(4) = \binom{4}{4}(0.9)^4(0.1)^0 = 1 \times 0.6561 \times 1 = 0.6561
Step 5: Add the two probabilities together.
P(k≥3)=0.2916+0.6561=0.9477P(k \geq 3) = 0.2916 + 0.6561 = 0.9477
Answer: The probability that at least 3 out of 4 bulbs work is approximately 0.948, or about 94.8%.

Frequently Asked Questions

What are the conditions for Bernoulli Trials?
Three conditions must hold: (1) each trial has exactly two outcomes — success or failure; (2) the probability of success p is the same on every trial; and (3) all trials are independent, meaning the outcome of one trial does not affect any other. If any condition is violated, you cannot use the binomial probability formula.
What is the difference between a Bernoulli Trial and a Binomial Distribution?
A single Bernoulli Trial is one experiment with two outcomes (success/failure). When you repeat that same Bernoulli Trial n independent times and count the total number of successes, the resulting probability distribution is called a binomial distribution. The binomial distribution is built from Bernoulli Trials.
When should you use the Bernoulli Trials formula?
Use it whenever you need the probability of getting exactly k successes in n identical, independent trials that each have only two outcomes. Common scenarios include coin flips, true/false quizzes, defective-item inspections, and free-throw shooting. If trials are not independent or the probability changes between trials, you need a different approach.

Bernoulli Trials (Binomial) vs. Geometric Distribution

Bernoulli Trials (Binomial)Geometric Distribution
Question askedWhat is the probability of exactly k successes in n trials?What is the probability that the first success occurs on trial number k?
Number of trialsFixed at nNot fixed — you keep going until the first success
FormulaP(k) = C(n,k) · p^k · q^(n−k)P(k) = q^(k−1) · p
ExampleProbability of 3 heads in 10 flipsProbability that the first head appears on the 4th flip

Why It Matters

Bernoulli Trials appear throughout statistics, from quality control in manufacturing to medical testing and sports analytics. In many standardized math courses (AP Statistics, IB Math, A-Level), binomial probability questions based on Bernoulli Trials are among the most heavily tested topics. Understanding them also lays the groundwork for more advanced distributions like the normal approximation to the binomial.

Common Mistakes

Mistake: Forgetting the binomial coefficient and only computing p^k · q^(n−k).
Correction: The term p^k · q^(n−k) gives the probability of one specific ordering of successes and failures. You must multiply by C(n, k) to account for all possible orderings of k successes among n trials.
Mistake: Applying the formula when trials are not independent.
Correction: If the outcome of one trial changes the probability for the next (e.g., drawing cards without replacement from a small deck), the trials are not independent and Bernoulli Trial assumptions are violated. In such cases, consider the hypergeometric distribution instead.

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