Skip to content
spmaddmath.com.my
Tuition

Study

SyllabusFormulasMethodsExam & PapersTools
LocationsPricingBlogOur TeachersContact
EN

Method · Probability Distribution

Using the binomial distribution

The binomial distribution gives the probability of exactly rr successes in nn independent trials with a constant success probability pp: P(X=r)=nCrpr(1p)nrP(X=r)={}^{n}C_{r}\,p^{r}(1-p)^{n-r}.

What the binomial distribution is for

The binomial distribution models a situation that is repeated a fixed number of times, where each repetition (a 'trial') has just two outcomes, a 'success' or a 'failure', and the probability of success stays the same every time. Tossing a coin 10 times and counting heads, or checking 5 items and counting how many are faulty, are classic examples.

If XX is the number of successes, then XX follows a binomial distribution written XB(n,p)X\sim B(n,p), where nn is the number of trials and pp is the probability of success in one trial. The formula P(X=r)=nCrpr(1p)nrP(X=r)={}^{n}C_{r}\,p^{r}(1-p)^{n-r} then gives the probability of exactly rr successes.

This lets us answer questions such as 'exactly 2', 'at least 3', or 'at most 1' success, and to find the mean npnp and variance np(1p)np(1-p).

When to reach for it

Reach for the binomial distribution when a question describes a fixed number of repeated trials and asks for the probability of a certain number of 'successes'. Look for four features: a set number of trials nn; each trial has only two outcomes; the trials are independent; and the success probability pp is the same each time.

Signal phrases include 'in nn trials', 'each with probability pp', 'exactly', 'at least', or 'at most'. If any trial affects the next, for example drawing without replacement so the probability changes, it is not binomial.

When the variable is instead a continuous measurement such as height or mass, you are looking at the normal distribution rather than the binomial.

The method, step by step

Binomial probabilityMust memorise
P(X=r)=nCrpr(1p)nrP(X=r)={}^{n}C_{r}\,p^{r}(1-p)^{n-r}
  1. 1

    Check the conditions

    Fixed nn trials, two outcomes per trial, independent trials, and a constant success probability pp.

  2. 2

    Identify n, p and r

    State the number of trials nn, the success probability pp, and the number of successes rr you need.

  3. 3

    Write the formula

    State P(X=r)=nCrpr(1p)nrP(X=r)={}^{n}C_{r}\,p^{r}(1-p)^{n-r}.

  4. 4

    Substitute the values

    Put nn, pp, 1p1-p and rr into the formula; check the two powers add up to nn.

  5. 5

    Evaluate

    Work out nCr{}^{n}C_{r}, then the two powers, then multiply the three parts together.

  6. 6

    Combine for 'at least' or 'at most'

    Add the separate terms you need, or use the complement, e.g. P(X1)=1P(X=0)P(X\ge 1)=1-P(X=0).

Worked example

Q1[4 marks]

In a certain game, the probability of winning any single round is 0.20.2. A player plays 5 independent rounds.

(a) Find the probability that the player wins exactly 2 rounds. (b) State the mean and variance of the number of rounds won.

Show worked solution

(a) The number of wins XX is binomial with n=5n=5 and p=0.2p=0.2, so 1p=0.81-p=0.8. We want P(X=2)P(X=2).

P(X=2)=5C2(0.2)2(0.8)3P(X=2)={}^{5}C_{2}(0.2)^{2}(0.8)^{3}

Evaluate each part: 5C2=10{}^{5}C_{2}=10, (0.2)2=0.04(0.2)^{2}=0.04, and (0.8)3=0.512(0.8)^{3}=0.512.

P(X=2)=10×0.04×0.512=0.2048P(X=2)=10\times 0.04\times 0.512=0.2048

The probability of winning exactly 2 rounds is 0.20480.2048.

(b) For a binomial distribution the mean is npnp and the variance is np(1p)np(1-p).

Mean=np=5×0.2=1,Variance=np(1p)=5×0.2×0.8=0.8\text{Mean}=np=5\times 0.2=1,\qquad \text{Variance}=np(1-p)=5\times 0.2\times 0.8=0.8

The mean number of rounds won is 11 and the variance is 0.80.8.

Common mistakes to avoid

  • Swapping pp and 1p1-p, or being unclear about which outcome counts as a 'success'.
  • Using the wrong rr, or writing exponents that do not add up to nn, the powers of pp and 1p1-p must total nn.
  • For 'at least one', calculating a long sum instead of using the complement 1P(X=0)1-P(X=0).
  • Confusing the mean npnp with the variance np(1p)np(1-p), or taking the variance as the standard deviation.
  • Rounding nCr{}^{n}C_{r} or the powers too early, so the final probability drifts off.

How one-to-one teaching helps

The step students most often get wrong is the very first one: checking that the situation really is binomial, and being clear about which outcome is the 'success' that pp belongs to. In a one-to-one lesson our teachers ask you to say the four conditions aloud and name nn, pp and rr before any calculating begins, so the setup is right every time.

We also drill the complement shortcut for 'at least' questions until it feels natural. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach the binomial distribution, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

What are the conditions for a binomial distribution?

There must be a fixed number of trials nn; each trial has only two outcomes (success or failure); the trials are independent; and the probability of success pp is constant across trials. If all four hold, the number of successes is binomial.

How do I find the mean and variance?

For XB(n,p)X\sim B(n,p), the mean is npnp and the variance is np(1p)np(1-p). The standard deviation is np(1p)\sqrt{np(1-p)}.

Remember the variance always uses the factor (1p)(1-p), so it is smaller than the mean.

How do I work out 'at least one' success?

It is usually fastest to use the complement: P(X1)=1P(X=0)P(X\ge 1)=1-P(X=0), where P(X=0)=nC0p0(1p)n=(1p)nP(X=0)={}^{n}C_{0}\,p^{0}(1-p)^{n}=(1-p)^{n}. This avoids adding many separate terms.

When is a situation not binomial?

When the trials are not independent or the success probability changes, for example, drawing objects without replacement, since each draw alters what is left. It is also not binomial when the variable is a continuous measurement like length or mass; that is modelled by the normal distribution.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Ready to get started?

Book a Trial Classfrom RM50/hr · One-hour paid trial · Same-day reply
Book a Trial ClassOne-hour paid trial · Same-day reply