# Binomial Distribution

See how to solve a binomial distribution:

finding P(X), E(X), V(X), σ(X).

3 examples and their solutions.

## Binomial Experiment

### Definition

B(n, p)

A binomial experiment isrepeating an independent event.

n: Number of repeating the event

p: Probability of the event

### Formula

P(X = k) =

P(X = k): Probability of the 'want' event happening k times. _{n}C_{k}⋅p^{k}⋅q^{n - k}_{n}C

_{k}: n trials, the number of ways to pick 'want' k times

q = 1 - p

(Probability of 'not want')

Binomial Theorem

Combination (Math)

### Example

A coin is tossed 7 times.

P(2 heads) = ?

Solution P(2 heads) = ?

B(7, 12) - [1]

q = 1 - 12

= 22 - 12

= 12 - [2]

P(X = 2) =

= 7⋅62⋅1⋅12

= 7⋅3⋅12

= 21128

q = 1 - 12

= 22 - 12

= 12 - [2]

P(X = 2) =

_{7}C_{2}⋅(12)^{2}⋅(12)^{5}- [3]= 7⋅62⋅1⋅12

^{2}⋅12^{5}= 7⋅3⋅12

^{7}- [4]= 21128

[1]

7 times

→ n = 7

Getting a head

→ p = 1/2

→ n = 7

Getting a head

→ p = 1/2

[2]

q: Not getting a head

→ Getting a tail

→ Getting a tail

[3]

7 trials,

the probability of picking a 'head' 2 times

the probability of picking a 'head' 2 times

Close

### Example

A fair die is tossed 3 times.

P(at most one '4') = ?

Solution P(at most one '4') = ?

B(3, 16) - [1]

q = 1 - 16

= 66 - 16

= 56 - [2]

P(X ≤ 1)

= P(X = 0) + P(X = 1) - [3]

=

= 1⋅1⋅5

= 5

= 5

= 5

= 5

= 5

= 2527

q = 1 - 16

= 66 - 16

= 56 - [2]

P(X ≤ 1)

= P(X = 0) + P(X = 1) - [3]

=

_{3}C_{0}⋅(16)^{0}⋅(56)^{3}+_{3}C_{1}⋅(16)^{1}⋅(56)^{2}= 1⋅1⋅5

^{3}6^{3}+ 3⋅16⋅5^{2}6^{2}= 5

^{3}+ 3⋅5^{2}6^{3}= 5

^{2}(5 + 3)6^{3}= 5

^{2}⋅8(2⋅3)^{3}= 5

^{2}⋅2^{3}2^{3}⋅3^{3}= 5

^{2}3^{3}= 2527

[1]

3 times

→ n = 3

Getting a '4'

→ p = 1/6

→ n = 3

Getting a '4'

→ p = 1/6

[2]

q: Not getting a '4'

[3]

(picking at most one '4')

= (picking no '4') + (picking one '4')

= (picking no '4') + (picking one '4')

Close

## Binomial Distribution

### Definition

B(n, p)

If n is big enough,the probability histogram shows

the shape of a binomial distribution.

Shape: Normal distribution

### Formula

B(n, p)

E(X) = np

V(X) = npq

σ(X) = √npq

E(X): Expected value E(X) = np

V(X) = npq

σ(X) = √npq

V(X): Variance

σ(X): Standard deviation

Variance, Standard Deviation

### Example

A fair die is tossed 180 times.

1. Find the expected value of the number of getting a '2'.

2. Find the variance of the number of getting a '2'.

3. Find the standard deviation of the number of getting a '2'.

Solution 1. Find the expected value of the number of getting a '2'.

2. Find the variance of the number of getting a '2'.

3. Find the standard deviation of the number of getting a '2'.

B(180, 16) - [1]

q = 1 - 16

= 66 - 16

= 56

1. E(X) = 180⋅16

= 30

2. V(X) = 180⋅16⋅56

= 30⋅56

= 25

3. σ(X) = √25

= 5

q = 1 - 16

= 66 - 16

= 56

1. E(X) = 180⋅16

= 30

2. V(X) = 180⋅16⋅56

= 30⋅56

= 25

3. σ(X) = √25

= 5

[1]

Close