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12. The Binomial Probability Distribution

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We use upper case variables (like X and Z) to denote random variables, and lower-case letters (like x and z) to denote specific values of those variables.

A binomial experiment is one that possesses the following properties:

  1. The experiment consists of n repeated trials;

  2. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);

  3. The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent.

The number of successes X in n trials of a binomial experiment is called a binomial random variable.

The probability distribution of the random variable X is called a binomial distribution, and is given by the formula:

`P(X)=C_x^n p^x q^(n-x)`


n = the number of trials

x = 0, 1, 2, ... n

p = the probability of success in a single trial

q = the probability of failure in a single trial

(i.e. q = 1 − p)

`C_x^n` is a combination

P(X) gives the probability of successes in n binomial trials.

Continues below

Mean and Variance of Binomial Distribution

If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. the mean value of the binomial distribution) is

E(X) = μ = np

The variance of the binomial distribution is

V(X) = σ2 = npq

Note: In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.

Example 1

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A die is tossed `3` times. What is the probability of

(a) No fives turning up?

(b) `1` five?

(c) `3` fives?


This is a binomial distribution because there are only `2` possible outcomes (we get a `5` or we don't).

Now, `n = 3` for each part. Let `X =` number of fives appearing.

(a) Here, x = 0.

`P(X=0)` `=C_x^np^xq^[n-x]` `=C_0^3 (1/6)^0 (5/6)^3` `=125/216` `=0.5787 `

(b) Here, x = 1.

`P(X=1)` `=C_x^np^xq^[n-x]` `=C_1^3 (1/6)^1 (5/6)^2` `=75/216` `=0.34722 `

(c) Here, x = 3.

`P(X=3)=C_x^np^xq^[n-x]` `=C_3^3 (1/6)^3 (5/6)^0` `=1/216` `=4.6296times10^-3 `

Example 2

Hospital records show that of patients suffering from a certain disease, `75%` die of it. What is the probability that of `6` randomly selected patients, `4` will recover?


This is a binomial distribution because there are only 2 outcomes (the patient dies, or does not).

Let X = number who recover.

Here, `n = 6` and `x = 4`. Let `p = 0.25` (success, that is, they live), `q = 0.75` (failure, i.e. they die).

The probability that `4` will recover:

`P(X)` `= C_x^np^xq^[n-x]` `=C_4^6(0.25)^4(0.75)^2` `=15times 2.1973 times 10^-3` `=0.0329595 `

Histogram of this distribution:

We could calculate all the probabilities involved and we would get:

`4``3.2959 times 10^-2`

The histogram is as follows:

It means that out of the `6` patients chosen, the probability that:

  • None of them will recover is `0.17798`,
  • One will recover is `0.35596`, and
  • All `6` will recover is extremely small.

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Example 3

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In the old days, there was a probability of `0.8` of success in any attempt to make a telephone call. (This often depended on the importance of the person making the call, or the operator's curiosity!)

Calculate the probability of having `7` successes in `10` attempts.


Probability of success `p = 0.8`, so `q = 0.2`.

`X =` success in getting through.

Probability of `7` successes in `10` attempts:





We use the following function


to obtain the probability histogram:

Example 4

A (blindfolded) marksman finds that on the average he hits the target `4` times out of `5`. If he fires `4` shots, what is the probability of

(a) more than `2` hits?

(b) at least `3` misses?


Here, `n = 4`, `p = 0.8`, `q = 0.2`.

Let `X =` number of hits.

Let x0 = no hits, x1 `= 1` hit, x2 `= 2` hits, etc.

(a) `P(X)=P(x_3)+P(x_4)`

`=C_3^4(0.8)^3(0.2)^1+` `C_4^4(0.8)^4(0.2)^0`



(b) `3` misses means `1` hit, and `4` misses means `0` hits.


`=C_1^4(0.8)^1(0.2)^3+` `C_0^4(0.8)^0(0.2)^4`



Example 5

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The ratio of boys to girls at birth in Singapore is quite high at `1.09:1`.

What proportion of Singapore families with exactly 6 children will have at least `3` boys? (Ignore the probability of multiple births.)

[Interesting and disturbing trivia: In most countries the ratio of boys to girls is about `1.04:1`, but in China it is `1.15:1`.]


The probability of getting a boy is `1.09/(1.09+1.00)=0.5215`

Let `X =` number of boys in the family.


`n = 6`,
`p = 0.5215`,
`q = 1 − 0.52153 = 0.4785`

When `x=3`:

` P(X)` `=C_x^np^xq^(n-x)` `=C_3^6(0.5215)^3(0.4785)^3` `=0.31077`

When `x=4`:

` P(X)` `=C_4^6(0.5215)^4(0.4785)^2` `=0.25402`

When `x=5`:

`P(X)` `=C_5^6(0.5215)^5(0.4785)^1` `=0.11074`

When `x=6`:

`P(X)` `=C_6^6(0.5215)^6(0.4785)^0` `=2.0115xx10^-2`

So the probability of getting at least 3 boys is:


`=0.31077+0.25402+` `0.11074+` `2.0115xx10^-2`


NOTE: We could have calculated it like this:

`P(X>=3)` `=1-(P(x_0)+P(x_1)+P(x_2))`

Easy to understand math videos:

Example 6

A manufacturer of metal pistons finds that on the average, `12%` of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of `10` pistons will contain

(a) no more than `2` rejects? (b) at least `2` rejects?


Let `X =` number of rejected pistons

(In this case, "success" means rejection!)

Here, `n = 10`, `p = 0.12`, `q = 0.88`.


No rejects. That is, when `x=0`:

`P(X)` `=C_x^np^xq^(n-x)` `=C_0^10(0.12)^0(0.88)^10` `=0.2785`

One reject. That is, when `x=1`

`P(X)` `=C_1^10(0.12)^1(0.88)^9` `=0.37977`

Two rejects. That is, when `x=2`:

`P(X)` `=C_2^10(0.12)^2(0.88)^8` `=0.23304`

So the probability of getting no more than 2 rejects is:


`=0.2785+` `0.37977+` `0.23304`


(b) We could work out all the cases for `X = 2, 3, 4, ..., 10`, but it is much easier to proceed as follows:

`"Probablity of at least 2 rejects"`


` =1-(P(x_0)+P(x_1))`

` =1-(0.2785+0.37977)`



Using the function `g(x)=C(10,x)(0.12)^x(0.88)^(10-x)` and finding the values at `0, 1, 2, ...`, gives us the histogram:


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