5.4 - The binomial distribution | Cambridge A-Level Mathematics

When the same chance experiment is repeated a fixed number of times, how can you find the chance of a particular number of successes? Learn when a binomial model fits, calculate exact and cumulative probabilities, and use its mean and variance to recover unknown parameters.

When a binomial model fits

Six independent attempts each have a success probability of $0.25$ . Is exactly two successes more likely than no successes? Listing every possible sequence would work, but a binomial model counts the relevant sequences for you.

First decide what one trial is and what you will call a success. Success is just the outcome being counted: a faulty component can be a success in the mathematical model.

Binomial distribution

The random variable $X$ has the binomial distribution $\mathrm B(n,p)$ when it counts successes in a fixed number $n$ of independent trials, each with two outcomes and the same success probability $p$ .

Write $X\sim\mathrm B(n,p)$ . Its possible values are $0,1,\ldots,n$ , where $n$ is a positive integer and $0\le p\le1$ .

An experiment can have several physical outcomes as long as they are grouped into success and failure. For a die, “a six” and “not a six” are two outcomes for this purpose. Independence means that knowing earlier outcomes does not change the chance of success on a later trial. Equal marginal probabilities alone do not establish independence.

Worked example 1: sampling with replacement

Question. A bag contains 3 red and 7 blue counters. A counter is drawn, its colour recorded and the counter replaced. The bag is mixed before each of 8 draws. State a distribution for the number $X$ of red counters drawn and justify it.

Plan. Check the trial count, the success definition, independence and the success probability.

There are exactly $8$ trials. Each is red or not red. Replacement restores the original contents, and the draws are modelled as independent after mixing. On every draw,

p=\frac3{10}=0.3.

Answer. $X\sim\mathrm B(8,0.3)$ .

Check. The count can be $0$ or $8$ , and every integer between them. These are precisely the possible values of this binomial variable.

Now change only the replacement rule. The first success probability stays the same, but later ones can change.

Worked example 2: sampling without replacement

Question. The same bag contains 3 red and 7 blue counters. Eight counters are drawn without replacement. Is $\mathrm B(8,0.3)$ an appropriate distribution for the number of reds?

Plan. Test whether an earlier outcome changes the next success probability.

After a red first draw, the probability of another red is $2/9$ . After a blue first draw it is $3/9$ . Neither conditional probability equals the original $3/10$ , so the trials are dependent.

Answer. No: the binomial independence and constant conditional success probability requirements fail.

Check. There are only $3$ red counters, so $X\le3$ . A $\mathrm B(8,0.3)$ model would assign positive probability to $X=8$ , an impossible outcome here.

In applications, a statement such as “independently, with probability $0.25$ on each attempt” supplies assumptions. If the probability changes through practice, fatigue or changing conditions, justify a binomial model rather than assuming it from a repeated experiment alone.

Watch out

“Two outcomes” is not enough. A fixed trial count, independence and the same success probability are also needed. Stopping as soon as the first success occurs does not give a fixed number of trials.

Choose the experiment whose conditions support the proposed model.

Exact and cumulative probabilities

One particular sequence with two successes in six attempts has probability $0.25^2\times0.75^4$ . It is not the only sequence: the two successes can occupy any $2$ of the $6$ positions, giving $\binom62=15$ sequences. Add their equal probabilities by multiplying by $15$ .

Binomial probabilities

For $X\sim\mathrm B(n,p)$ , where $0<p<1$ , and an integer $r$ with $0\le r\le n$ ,

\mathrm P(X=r)=\binom nr p^r(1-p)^{n-r}.

In the formula booklet.

The powers count successes and failures. The coefficient counts the positions of the successes. At the endpoint cases, $p=0$ gives $X=0$ with certainty and $p=1$ gives $X=n$ with certainty. For “at most”, add the probabilities of the permitted integer values. For “at least”, a complement often uses fewer terms.

Worked example 3: exactly two successes

Question. Six independent attempts each have success probability $0.25$ . Find the probability of exactly two successes.

Plan. Use $n=6$ , $p=0.25$ and $r=2$ in the binomial formula.

\begin{aligned} \mathrm P(X=2)&=\binom62(0.25)^2(0.75)^4\\ &=15\times\frac1{16}\times\frac{81}{256}\\ &=0.296630859375 \end{aligned}

Answer. $0.297$ , to $3$ significant figures.

Check. Every two-success sequence has probability $81/4096$ . Multiplying by the $15$ possible pairs of success positions gives $1215/4096$ , the same result. Omitting the coefficient would count just one sequence.

Change “exactly two” to “at least two”. The model is unchanged; the event now includes $2,3,4,5,6$ .

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5.4 - The binomial distribution

When a binomial model fits

Exact and cumulative probabilities

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