# 12. The Binomial Probability Distribution

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### Notation

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:

The experiment consists of

*n*repeated trials;Each trial results in an outcome that may be classified as a

**success**or a**failure**(hence the name,**binomial**);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)`

where

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.

## 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) =σ^{2}=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?

Answer

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 `

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### 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?

Answer

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:

`X` | `text[Probability]` |

`0` | `0.17798` |

`1` | `0.35596` |

`2` | `0.29663` |

`3` | `0.13184` |

`4` | `3.2959 times 10^-2` |

`5` | `4.3945times10^-3` |

`6` | `2.4414times10^-4` |

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.

### 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.

Answer

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

`X =` success in getting through.

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

`text[Probability]=P(X=7)`

`=C_7^10(0.8)^7(0.2)^[10-7]`

`=0.20133`

#### Histogram

We use the following function

`C(10,x)(0.8)^x(0.2)^[10-x]`

to obtain the probability histogram:

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### 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?

Answer

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

Let `X =` number of hits.

Let *x*_{0} = no hits, *x*_{1} `= 1` hit, *x*_{2} `= 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`

`=4(0.8)^3(0.2)+(0.8)^4`

`=0.8192`

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

`P(X)=P(x_1)+P(x_0)`

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

`=4(0.8)^1(0.2)^3+(0.2)^4`

`=0.0272`

### 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`.]

Answer

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

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

Here,

`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:

`"Probability"=P(X>=3)`

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

`=0.69565`

**NOTE:** We could have calculated it like this:

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

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### 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?

Answer

Let `X =` number of rejected pistons

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

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

(a)

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:

`"Probability"=P(X<=2)`

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

`=0.89131`

(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<=1)`

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

` =1-(0.2785+0.37977)`

`=0.34173`

#### Histogram

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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