# 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) = nCkpkqn - 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

### 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(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, 7C2
times,
p, 1/2 to the 2
times,
q, 1/2 to the 7 - 2 = 5.

7C2 = [7⋅6]/[2⋅1]

Combination

[1/2]2 = 1/22
[1/2]5 = 1/25

Power of a Quotient

Cancel the denominator 2⋅1
and reduce 6 in the numerator to, 6/2, 3.

[1/22]⋅[1/25]
= 1/22 + 5
= 1/27

7⋅3 = 21

27 = 128

## Example 2

### 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, 3C0
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, 3C1
times,
p, 1/6 to the 1
times,
q, 5/6 to the 3 - 1 = 2.

So
P(X = 0) + P(X = 1)
= 3C0⋅[1/6]0⋅[5/6]3 + 3C1⋅[1/6]1⋅[5/6]2.

3C0 = 1
[1/6]0 = 1
[5/6]3 = 53/63

3C1 = 3/1 = 3
[1/6]1 = 1/6
[5/6]2 = 52/62

1⋅1⋅53 = 53
+3⋅1⋅52 = +3⋅25

+3⋅25 = +75

125 + 75 = 200

200 = 23⋅52
63 = 23⋅33

Prime Factorization

Cancel 23 factors.

52 = 25
33 = 27