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Binomial Probability Formula

Binomial Probability Formula

A probability formula for Bernoulli trials. The probability of achieving exactly k successes in n trials is shown below.

Formula:

P(k successes in n trials) = (n choose k) times p^k times 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 out of 10) = C(10,7) × (0.25)^7 × (0.75)^3 ≈ 0.0031

See also

Binomial coefficients

Key Formula

P(X=k)=(nk) pk qn−kP(X = k) = \binom{n}{k} \, p^k \, q^{n-k}
Where:
  • nn = Total number of independent trials
  • kk = Number of successes you want
  • pp = Probability of success on a single trial
  • qq = Probability of failure on a single trial, where q = 1 − p
  • (nk)\binom{n}{k} = Binomial coefficient, equal to n! / (k!(n−k)!), representing the number of ways to choose k successes from n trials

Worked Example

Problem: You flip a fair coin 8 times. What is the probability of getting exactly 5 heads?
Step 1: Identify the variables. Each flip is an independent trial with two outcomes. Here n = 8, k = 5, p = 0.5 (probability of heads), and q = 0.5 (probability of tails).
n=8,k=5,p=0.5,q=0.5n = 8, \quad k = 5, \quad p = 0.5, \quad q = 0.5
Step 2: Calculate the binomial coefficient. This counts the number of ways to choose which 5 of the 8 flips land heads.
(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 any single arrangement with exactly 5 heads and 3 tails.
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-arrangement probability.
P(X=5)=56×1256=56256=732≈0.2188P(X = 5) = 56 \times \frac{1}{256} = \frac{56}{256} = \frac{7}{32} \approx 0.2188
Answer: The probability of getting exactly 5 heads in 8 flips is 7/32, or approximately 0.2188 (about 21.9%).

Another Example

This example uses an unequal probability (p ≠ 0.5), showing how the formula handles a situation where success and failure are not equally likely. It also uses a real-world sports scenario rather than coin flips.

Problem: A basketball player makes 80% of her free throws. In a game, she shoots 6 free throws. What is the probability she makes exactly 4 of them?
Step 1: Identify the variables. Each free throw is independent with a constant success rate. Here n = 6, k = 4, p = 0.8, and q = 0.2.
n=6,k=4,p=0.8,q=0.2n = 6, \quad k = 4, \quad p = 0.8, \quad q = 0.2
Step 2: Calculate the binomial coefficient.
(64)=6!4!⋅2!=6×52×1=15\binom{6}{4} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15
Step 3: Compute the success and failure probability terms separately.
p4=(0.8)4=0.4096,q2=(0.2)2=0.04p^4 = (0.8)^4 = 0.4096, \quad q^2 = (0.2)^2 = 0.04
Step 4: Multiply all three components together.
P(X=4)=15×0.4096×0.04=15×0.016384=0.24576P(X = 4) = 15 \times 0.4096 \times 0.04 = 15 \times 0.016384 = 0.24576
Answer: The probability she makes exactly 4 out of 6 free throws is approximately 0.2458, or about 24.6%.

Frequently Asked Questions

When do you use the binomial probability formula?
You use it when you have a fixed number of independent trials, each trial has exactly two outcomes (success or failure), and the probability of success stays the same from trial to trial. Common examples include coin flips, multiple-choice guessing, quality control inspections, and free throw shooting.
What is the difference between binomial probability and normal probability?
Binomial probability applies to discrete outcomes — you count exact whole numbers of successes. Normal probability uses a continuous bell curve and calculates probabilities for ranges of values. When n is large, the binomial distribution can be approximated by a normal distribution, which is a useful shortcut.
How do you find the probability of getting 'at least' k successes using the binomial formula?
To find P(X ≥ k), you sum the binomial probabilities for every value from k up to n: P(X = k) + P(X = k+1) + … + P(X = n). Alternatively, you can use the complement: P(X ≥ k) = 1 − P(X ≤ k−1). The complement method often requires fewer calculations.

Binomial Probability vs. Geometric Probability

Binomial ProbabilityGeometric Probability
Question answeredProbability of exactly k successes in n trialsProbability that the first success occurs on the kth trial
Number of trialsFixed at nNot fixed — you keep going until the first success
FormulaP(X = k) = C(n,k) · p^k · q^(n−k)P(X = k) = q^(k−1) · p
Typical use caseHow many heads in 10 flips?How many flips until the first head?

Why It Matters

The binomial probability formula appears throughout high school statistics, AP Statistics, and introductory college courses. You will use it to solve problems about quality control, medical testing, survey sampling, and games of chance. Understanding this formula also lays the groundwork for the normal approximation to the binomial and for more advanced distributions like the negative binomial and the Poisson.

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 arrangement of successes and failures. You must multiply by C(n, k) to count all the different ways those k successes can be distributed among the n trials.
Mistake: Swapping the exponents — putting (n − k) on p and k on q.
Correction: The exponent on p must equal the number of successes k, and the exponent on q must equal the number of failures n − k. A quick check: the two exponents should add up to n.

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