# Binomial Experiment

How to find the probability of a binomial experiment: definition, formula, 2 examples, and their solutions.

## Definition

### Definition

A binomial experiment is

repeating an independent event.

If an event whose probability is p

repeats n times,

then the binomial experiment can be written as

B(n, p).

## Formula

### Formula

For a binomial experiment B(n, p),

the probability of the wanted event

happening k times is

P(X = k) = _{n}C_{k}p^{k}q^{n - k}.

n: Number of the binomial experiment

k: Number of the wanted event happening

p: Probability of the wanted event

q = 1 - p

Probability: Not A

The formula looks like

a term of the binomial theorem.

## Example 1

### Example

### Solution

First, write the given condition as

B(n, p).

Getting a head of a coin

isn't affected by the previous trial.

So this is an independent event.

This is repeated.

So this is a binomial experiment.

n = 7

So write

B(7.

The probability of getting a head of a coin is

1/2.

So p = 1/2.

So write

1/2).

So the given binomial experiment is

B(7, 1/2).

p = 1/2

So q = 1 - 1/2 = 1/2.

P(2 heads) is

P(X = 2).

B(7, 1/2)

n = 7

p = 1/2

q = 1/2

Then P(X = 2) is,

pick 2 of p from 7 trials, _{7}C_{2}

times,

p, 1/2 to the 2

times,

q, 1/2 to the 7 - 2 = 5.

_{7}C_{2} = [7⋅6]/[2⋅1]

Combination

[1/2]^{2} = 1/2^{2}

[1/2]^{5} = 1/2^{5}

Power of a Quotient

Cancel the denominator 2⋅1

and reduce 6 in the numerator to, 6/2, 3.

[1/2^{2}]⋅[1/2^{5}]

= 1/2^{2 + 5}

= 1/2^{7}

7⋅3 = 21

2^{7} = 128

So 21/128 is the answer.

## Example 2

### Example

### Solution

First, write the given condition as

B(n, p).

Getting a '4' of a fair die

isn't affected by the previous trial.

So this is an independent event.

This is repeated.

So this is a binomial experiment.

n = 3

So write

B(3.

The probability of getting a '4' of a die is

1/6.

So p = 1/6.

So write

1/6).

So the given binomial experiment is

B(3, 1/6).

p = 1/6

So q = 1 - 1/6 = 5/6.

P(at most one '4') is

P(X ≤ 1).

X ≤ 1 means

x = 0: getting no '4'

or

X = 1: getting one '4'.

(X is the number of the wanted events.)

So

P(X ≤ 1) = P(X = 0) + P(X = 1).

Find P(X = 0).

B(3, 1/6)

n = 3

p = 1/6

q = 5/6

Then P(X = 0) is,

pick 0 of p from 3 trials, _{3}C_{0}

times,

p, 1/6 to the 0

times,

q, 5/6 to the 3 - 0 = 3.

Find P(X = 1).

P(X = 1) is,

pick 1 of p from 3 trials, _{3}C_{1}

times,

p, 1/6 to the 1

times,

q, 5/6 to the 3 - 1 = 2.

So

P(X = 0) + P(X = 1)

= _{3}C_{0}⋅[1/6]^{0}⋅[5/6]^{3} + _{3}C_{1}⋅[1/6]^{1}⋅[5/6]^{2}.

_{3}C_{0} = 1

[1/6]^{0} = 1

[5/6]^{3} = 5^{3}/6^{3}_{3}C_{1} = 3/1 = 3

[1/6]^{1} = 1/6

[5/6]^{2} = 5^{2}/6^{2}

1⋅1⋅5^{3} = 5^{3}

+3⋅1⋅5^{2} = +3⋅25

+3⋅25 = +75

125 + 75 = 200

200 = 2^{3}⋅5^{2}

6^{3} = 2^{3}⋅3^{3}

Prime Factorization

Cancel 2^{3} factors.

5^{2} = 25

3^{3} = 27

So 25/27 is the answer.