# Binomial Experiment

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

## Formula

Think about an independent event

whose P(want) is *p*.

And set P(not want) as *q*. (= 1 - *p*)

Probability of (not *A*, Complement)

For *n* trials,

if the [want] happens [*k*] times,

then [not want] happens [*n* - *k*] times.

It's like finding the ways of

choosing *p* [*k*] times

and choosing *q* [*n* - *k*] times.

Then, by the binomial theorem,

the probability of [want] happening *k* times,

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

(*q* = 1 - *p*)

Binomial theorem

## Example 1

The result of the past tossing

does not affect the next tossing.

So tossing a coin is an independent event.

And the same tossing is repeated.

So this is a binomial experiment problem.

A coin is tossed 6 times.

So *n* = 6.

It says to find P(3 heads).

So *k* = 3.

For a single toss,

the probability of getting a head is 1/2.

So *p* = 1/2.

*p* = 1/2

So *q* = 1 - 1/2

= 1/2.

This *q* means

the probability of not getting a head:

getting a tail.

Probability of (not *A*, Complement)

*n* = 6*k* = 3*p* = 1/2*q* = 1/2

So P(*X* = 3) is,

choose *p* 3 times and choose *q* 6 - 3 = 3 times,_{6}C_{3}⋅(1/2)^{3}⋅(1/2)^{3}.

_{6}C_{3} = (6⋅5⋅4)/(3⋅2⋅1)

Combinations (_{n}C_{r})

(1/2)^{3} = 1/8

Cancel the numerator 6

and cancel 2⋅3.

Cancel the numerator 4

and reduce the denominator 8 to 2.

5⋅(1/2)⋅(1/8) = 5/16

So the probability is 5/16.

## Example 2

The result of the past spinning

does not affect the next spinning.

So spinning a spinner is an independent event.

And the same spinning is repeated.

So this is a binomial experiment problem.

A spinner is spun 5 times.

So *n* = 5.

It says to find P(2 'blue's).

So *k* = 2.

The central angle of the blue part is 120º.

And the central angle of the whole spinner is 360º.

So *p* = 120/360.

So *p* = 1/3.

*p* = 1/3

So *q* = 1 - 1/3

= 2/3.

*n* = 5*k* = 2*p* = 1/3*q* = 2/3

So P(*X* = 2) is,

choose *p* 2 times and choose *q* 5 - 2 = 3 times,_{5}C_{2}⋅(1/3)^{2}⋅(2/3)^{3}.

_{5}C_{2} = (5⋅4)/(2⋅1)

Combinations (_{n}C_{r})

(1/3)^{2} = 1/9

(2/3)^{3} = 8/27

Cancel the denominator 2

and reduce the numerator 4 to 2.

9⋅27 = 243

5⋅2 = 10

So 5⋅2⋅8 = 80

So the probability is 80/243.

## Example 3

The result of the past tossing

does not affect the next tossing.

So tossing a die is an independent event.

And the same tossing is repeated.

So this is a binomial experiment problem.

A coin is tossed 3 times.

So *n* = 3.

It says to find P(at most one '6').

This includes [no '6'] or [one '6'].

So *k* = 0 and 1.

For a single toss,

the probability of getting a '6' is 1/6.

So *p* = 1/6.

*p* = 1/6

So *q* = 1 - 1/6

= 5/6.

This *q* means

the probability of not getting a '6'.

P(at most one '6') means P(*X* ≤ 1).

This includes P(*X* = 0) and P(*X* = 1).

So P(*X* ≤ 1) = P(*X* = 0) + P(*X* = 1).

*n* = 3*k* = 0, 1*p* = 1/6*q* = 5/6

So P(*X* = 0) is,

choose *p* 3 times and choose *q* 3 - 0 = 3 times,_{3}C_{0}⋅(1/6)^{0}⋅(5/6)^{3}.

And P(*X* = 1) is,

choose *p* 1 time and choose *q* 3 - 1 = 2 times,_{3}C_{1}⋅(1/6)^{1}⋅(5/6)^{2}.

_{3}C_{0} = 1

(5/6)^{3} = 5^{3}/6^{3}_{3}C_{1} = 3

(5/6)^{2} = 25/6^{2}

5^{3} = 125

3⋅25 = 75

So the probability is

125/6^{3} + 75/6^{3}.

125 + 75 = 200

To reduce 200 and 6^{3},

change them as their prime factorizations.

200 = 2^{3}⋅5^{2}

6^{3} = 2^{3}⋅3^{3}

Prime factorization

Cancel the common factors 2^{3}.

5^{2} = 25

3^{3} = 27

Then the probability is 25/27.