The Binomial Distribution with Examples

The binomial distribution models one of the most common situations in probability: repeating the same yes-or-no experiment a fixed number of times and counting the successes. How many heads in ten coin tosses? How many correct answers if you guess on a multiple-choice test? This guide explains when to use the binomial distribution and how to calculate its probabilities with confidence.

When Is a Situation Binomial?

A random variable \( X \) follows a binomial distribution if all four of these conditions hold:

1. Fixed number of trials, \( n \).
2. Two outcomes per trial, called "success" and "failure".
3. Constant probability of success, \( p \), on every trial.
4. Independent trials: one result does not affect another.

We then write

$$ X \sim B(n, p) $$

where \( X \) is the number of successes. The probability of failure is \( q = 1 - p \).

A quick way to remember the conditions is the acronym BINS: Binary outcomes, Independent trials, Number of trials fixed, Same probability.

Examples and Non-Examples

• Tossing a coin 10 times and counting heads: binomial, \( B(10, 0.5) \).
• Drawing 5 cards without replacement and counting hearts: not binomial, because the probability changes after each draw.
• Rolling a die until you get a six: not binomial, because the number of trials is not fixed.

The Binomial Probability Formula

$$ P(X = r) = \binom{n}{r}\, p^{r}\, q^{\,n-r}, \qquad r = 0, 1, 2, \dots, n $$

where

$$ \binom{n}{r} = \frac{n!}{r!\,(n-r)!} $$

counts the number of ways to arrange \( r \) successes among \( n \) trials.

Why it works: any single sequence with \( r \) successes and \( n - r \) failures has probability \( p^r q^{n-r} \), and there are \( \binom{n}{r} \) such sequences.

Mean, Variance and Standard Deviation

For \( X \sim B(n, p) \):

$$ \mu = E(X) = np, \qquad \sigma^2 = \operatorname{Var}(X) = npq, \qquad \sigma = \sqrt{npq} $$

These formulas save a great deal of time compared with calculating from the full distribution.

Worked Example 1: Customers Making a Purchase

Question: In a shop, 30% of customers who enter make a purchase. If 10 customers enter independently, find the probability that exactly 3 make a purchase.

Step 1: Identify the distribution. \( n = 10 \), \( p = 0.3 \), \( q = 0.7 \), so \( X \sim B(10, 0.3) \).

Step 2: Compute the combination.

$$ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 $$

Step 3: Compute the powers.

$$ 0.3^3 = 0.027, \qquad 0.7^7 = 0.0823543 $$

Step 4: Multiply.

$$ P(X = 3) = 120 \times 0.027 \times 0.0823543 \approx 0.2668 $$

There is about a 26.7% chance that exactly 3 of the 10 customers buy something.

Mean and standard deviation: \( \mu = 10 \times 0.3 = 3 \) and \( \sigma = \sqrt{10 \times 0.3 \times 0.7} = \sqrt{2.1} \approx 1.449 \). Notice that 3 is also the expected number, which is why it is one of the most likely values.

Worked Example 2: Guessing on a Multiple-Choice Test

Question: A quiz has 5 multiple-choice questions, each with 4 options. A student guesses every answer. Find (a) the probability of getting at least 3 correct, (b) the mean and standard deviation of the number correct.

Step 1: Identify the distribution. \( n = 5 \), \( p = \frac{1}{4} \), \( q = \frac{3}{4} \), so \( X \sim B\left(5, \tfrac{1}{4}\right) \).

Step 2: Break "at least 3" into cases.

$$ P(X \ge 3) = P(X = 3) + P(X = 4) + P(X = 5) $$

Step 3: Each term.

$$ P(X = 3) = \binom{5}{3}\left(\tfrac{1}{4}\right)^3\left(\tfrac{3}{4}\right)^2 = 10 \times \frac{1}{64} \times \frac{9}{16} = \frac{90}{1024} $$

$$ P(X = 4) = \binom{5}{4}\left(\tfrac{1}{4}\right)^4\left(\tfrac{3}{4}\right) = 5 \times \frac{1}{256} \times \frac{3}{4} = \frac{15}{1024} $$

$$ P(X = 5) = \left(\tfrac{1}{4}\right)^5 = \frac{1}{1024} $$

Step 4: Add.

$$ P(X \ge 3) = \frac{90 + 15 + 1}{1024} = \frac{106}{1024} \approx 0.1035 $$

Guessing gives only about a 10% chance of passing with 3 or more correct.

Step 5: Mean and standard deviation.

$$ \mu = 5 \times \tfrac{1}{4} = 1.25, \qquad \sigma^2 = 5 \times \tfrac{1}{4} \times \tfrac{3}{4} = 0.9375, \qquad \sigma \approx 0.968 $$

On average, a guessing student gets 1.25 questions right.

Cumulative Probabilities and the Complement

Questions often use phrases like "at most", "fewer than" or "at least". Translate them carefully:

• "At most 2" means \( P(X \le 2) \).
• "Fewer than 2" means \( P(X \le 1) \).
• "At least 1" means \( P(X \ge 1) = 1 - P(X = 0) \).

For example, in Example 2, the probability of getting at least one answer right is

$$ 1 - \left(\tfrac{3}{4}\right)^5 = 1 - \frac{243}{1024} = \frac{781}{1024} \approx 0.763 $$

Common Mistakes

• Using the formula when the conditions fail, such as sampling without replacement from a small group.
• Forgetting the combination \( \binom{n}{r} \) and calculating only one sequence.
• Misreading inequalities: "more than 3" is \( X \ge 4 \), not \( X \ge 3 \).
• Mixing up \( p \) and \( q \) when the "success" is defined as something bad, like a defect.
• Using \( np \) as the standard deviation. The standard deviation is \( \sqrt{npq} \).

Tips

• Define success clearly in words before writing \( p \).
• Use the complement whenever the direct sum has many terms.
• Learn your calculator's binomial PD and CD functions to check answers quickly.

Key Takeaways

• Use \( B(n, p) \) for a fixed number of independent trials with two outcomes and constant \( p \).
• \( P(X = r) = \binom{n}{r} p^r q^{n-r} \).
• Mean \( np \), variance \( npq \), standard deviation \( \sqrt{npq} \).
• Break "at least" and "at most" questions into cases or use the complement.

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