Exactly r Successes: How to Use the Binomial Probability Formula

With n independent trials, two outcomes per trial and a constant success probability p, the probability of exactly r successes is C(n,r)p^r(1 − p)^(n − r).

Check the experiment before choosing a formula

The number of trials must be fixed, and a success must mean the same thing each time. “Success” is simply the outcome being counted, not necessarily a favourable result. Independent trials do not influence each other’s outcome probabilities. If items are sampled without replacement from a small collection, these conditions generally fail.

The integer r must lie between 0 and n inclusive. The combination C(n,r) counts which r trial positions are successes. The powers calculate the probability of any one arrangement. This separation between counting arrangements and assigning probabilities explains the formula.

Worked example: a fair coin

Find the probability of exactly two heads in five independent tosses of a fair coin. Here n = 5, r = 2 and p = 1/2. One particular arrangement, HHTTT, has probability (1/2)²(1/2)³ = 1/32. There are C(5,2) = 5 × 4/(2 × 1) = 10 positions for the two heads.

Therefore P(X = 2) = 10/32 = 5/16 = 0.3125. This includes arrangements such as HTHTT and TTTHH, not just the first arrangement you write down. The event says exactly two heads, so three, four and five heads are excluded.

A non-fair example

Suppose a model experiment has three independent trials, each with success probability 0.2. Exactly one success has probability C(3,1)(0.2)(0.8)² = 3 × 0.2 × 0.64 = 0.384. Unlike a fair coin, the eight success/failure strings are not equally likely. Their probabilities depend on the numbers of successes and failures.

For “at most r,” add the probabilities from 0 through r. For “at least one,” the complement 1 − P(X = 0) is often shorter. Do not replace these events by the exactly-r formula without changing the calculation.

Common mistake

Omitting C(n,r) gives only one arrangement’s probability. Another error is using p for both success and failure when p is not 1/2. Write q = 1 − p, then check that the exponents total n. Finally, a probability is a prediction under a model, not a promise that a short experiment will reproduce that proportion.

Exactly kitne success, aur unki kitni positions—dono gino.

Try three questions, then check the working

1. What is the probability of exactly three heads in four fair independent tosses?

   C(4,3) = 4. Each four-toss arrangement has probability 1/16. Multiply to obtain 4/16 = 1/4.

2. Two independent trials have p = 0.3. Find the chance of zero successes.

   Use C(2,0)(0.3)⁰(0.7)² = 1 × 1 × 0.49 = 0.49. Zero successes means both trials fail.

3. Three independent trials have p = 0.4. Find the chance of exactly two successes.

   C(3,2)(0.4)²(0.6) = 3 × 0.16 × 0.6 = 0.288. The three arrangements have the same probability.

Build the next step

• Probability Class 10 Notes: Formula, Sample Space, Coins, Dice and Cards
• Probability Class 12 Notes: Conditional Probability, Bayes’ Theorem and Distributions

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Reference and authorship

The examples and practice questions here are original, with calculations checked separately. Definitions and related background: OpenStax: Binomial Distribution. Prepared with AI assistance and editorial arithmetic review; no textbook affiliation is claimed.