What is the probability of observing exactly k successes in n trials in a…

Concept

The binomial distribution models the number of successes in n independent trials, each with a constant success probability p. The probability of observing exactly k successes equals the probability of one fixed ordering of k successes and (n − k) failures, multiplied by the number of distinct orderings in which those k successes can occur.

Application

1. Probability of one fixed ordering: because trials are independent, one specific sequence with exactly k successes (each with probability p) and (n − k) failures (each with probability 1 − p) has probability pk(1 − p)n−k.

2. Count the orderings: the k successes can occupy any k of the n trial positions, and the number of ways to choose those positions is the combination C(n, k) = n! / (k!(n − k)!).

3. Combine: multiplying the probability of one ordering by the number of equally likely orderings gives the probability of exactly k successes: P(X = k) = C(n, k)·pk·(1 − p)n−k.

Cross-check

Setting n = 1 makes C(1, k) equal to 1 for k = 0 or k = 1, so the formula reduces to the plain single-trial probabilities (1 − p) and p — consistent with a binomial distribution being n repeated Bernoulli trials. Summing P(X = k) over k = 0 to n also reproduces the binomial expansion of (p + (1 − p))n = 1, confirming the probabilities across all k values add up correctly.

Why the other forms fall short

• p(1 − p)k has no combination factor at all and does not even track the total number of trials n in its exponents.

• pk(1 − p)n−k correctly captures one ordering but has no combination factor, so it never counts how many orderings of k successes among n trials exist.

• nCk·pn(1 − p)k keeps a combination factor but the two exponents are interchanged, so they no longer track which count (successes or failures) belongs to which base.

Result: P(X = k) = C(n, k)·pk·(1 − p)n−k — the combination term multiplied by p raised to the successes count and (1 − p) raised to the failures count.