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) = nCk⋅pk⋅qn - k
P(X = k): Probability of the 'want' event happening k times
nCk: 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) = 7C2⋅(12)2⋅(12)5 - [3]
= 7⋅62⋅1⋅122⋅125
= 7⋅3⋅127 - [4]
= 21128
q = 1 - 12
= 22 - 12
= 12 - [2]
P(X = 2) = 7C2⋅(12)2⋅(12)5 - [3]
= 7⋅62⋅1⋅122⋅125
= 7⋅3⋅127 - [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]
= 3C0⋅(16)0⋅(56)3 + 3C1⋅(16)1⋅(56)2
= 1⋅1⋅5363 + 3⋅16⋅5262
= 53 + 3⋅5263
= 52(5 + 3)63
= 52⋅8(2⋅3)3
= 52⋅2323⋅33
= 5233
= 2527
q = 1 - 16
= 66 - 16
= 56 - [2]
P(X ≤ 1)
= P(X = 0) + P(X = 1) - [3]
= 3C0⋅(16)0⋅(56)3 + 3C1⋅(16)1⋅(56)2
= 1⋅1⋅5363 + 3⋅16⋅5262
= 53 + 3⋅5263
= 52(5 + 3)63
= 52⋅8(2⋅3)3
= 52⋅2323⋅33
= 5233
= 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