What if k is greater than n?

The probability is strictly 0. You cannot have more successes than the total number of trials.

How do you handle 'at least' or 'at most' probability questions?

Sum the individual binomial probabilities for all relevant values of $k$. For example, $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. Alternatively, use the complement rule: $P(X \ge 1) = 1 - P(X=0)$.

Can I use this formula if subjects are selected without replacement?

No. If items are drawn without replacement, the probability of success changes with each trial. Use the hypergeometric distribution instead.

How to calculate a binomial probability

The binomial probability formula calculates the chance of observing exactly $k$ successes in $n$ independent trials.

It applies only when there is a fixed number of trials, each with exactly two possible outcomes, and the probability of success remains constant across every trial.

The setup

Identify the four required parameters: $n$ (the total number of trials), $k$ (the target number of successes), $p$ (the probability of success on a single trial), and $1-p$ (the probability of failure). Verify that the trials are strictly independent.

The steps

Calculate the binomial coefficient, which represents the number of ways to choose $k$ successes from $n$ trials: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ .
Compute the probability of the required successes: $p^k$ .
Compute the probability of the required failures: $(1-p)^{n-k}$ .
Multiply the components to find the total probability: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ .

Checking the result

Verify that the final calculated probability satisfies $0 \le P(X=k) \le 1$ . If you calculate $P(X=k)$ for all possible values of $k$ from $0$ to $n$ , their total sum must equal exactly $1$ .

Common errors

A frequent mistake is swapping the values of $p$ and $1-p$ , which evaluates the wrong outcome. Another common error is applying this formula to dependent trials, such as drawing cards from a deck without replacement.

Worked example

A biased coin has a 0.6 probability of landing on heads. If the coin is flipped 5 times, what is the probability it lands on heads exactly 3 times?

Identify parameters: $n = 5$ , $k = 3$ , $p = 0.6$ . Calculate the binomial coefficient: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{120}{6 imes 2} = 10$ . Compute success probability: $p^k = 0.6^3 = 0.216$ . Compute failure probability: $(1-p)^{n-k} = (1-0.6)^{5-3} = 0.4^2 = 0.16$ . Multiply the terms: $P(X=3) = 10 imes 0.216 imes 0.16 = 0.3456$ .

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References: OpenStax Introductory Statistics, Chapter 4: Discrete Random Variables · Khan Academy, Statistics and Probability: Binomial probability formula