Binomial Experiment

Binomial Experiment

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

Formula

P(X = k) = n_C_k * p^k * q^(n - k). n: Number of trials, k: Number of [want] happened, p: Probability of [want], q = 1 - p.

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,
nCkpkqn - k.
(q = 1 - p)

Binomial theorem

Example 1

A coin is tossed 6 times. Find P(3 heads).

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,
6C3⋅(1/2)3⋅(1/2)3.

6C3 = (6⋅5⋅4)/(3⋅2⋅1)

Combinations (nCr)

(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

For the given spinner, when spinning it 5 times, find P(2 'blue's).

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,
5C2⋅(1/3)2⋅(2/3)3.

5C2 = (5⋅4)/(2⋅1)

Combinations (nCr)

(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

A fair die is tossed 3 times. Find P(at most one '6').

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,
3C0⋅(1/6)0⋅(5/6)3.

And P(X = 1) is,
choose p 1 time and choose q 3 - 1 = 2 times,
3C1⋅(1/6)1⋅(5/6)2.

3C0 = 1
(5/6)3 = 53/63

3C1 = 3
(5/6)2 = 25/62

53 = 125

3⋅25 = 75

So the probability is
125/63 + 75/63.

125 + 75 = 200

To reduce 200 and 63,
change them as their prime factorizations.

200 = 23⋅52

63 = 23⋅33

Prime factorization

Cancel the common factors 23.

52 = 25
33 = 27

Then the probability is 25/27.