Binomial Distribution

First, write the given condition as
B(n, p).

Getting a '2' 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 = 180

So write
B(180.

The probability of getting a '2' of a die is
1/6.

So p = 1/6.

So write
1/6).

So the given binomial experiment is
B(180, 1/6).

B(180, 1/6)
n = 180
p = 1/6

Then the expected value E(X) is
180⋅[1/6].

180⋅[1/6] = 30

So E(X) = 30.

You just found that
B(180, 1/6).

B(180, 1/6)
p = 1/6

Then q = 1 - 1/6 = 5/6.

B(180, 1/6)
n = 180
p = 1/6

q = 5/6

Then the variance V(X) is
180⋅[1/6]⋅[5/6].

180⋅[1/6] = 30

Cancel the denominator 6
and reduce 30 to, 30/6, 5.

5⋅5 = 25

So V(X) = 25.

The condition is different
from the previous examples.

So write the given condition as
B(n, p).

The probability of
getting 'the multiple of 3', 3 or 6, is
2/6 = 1/3.

So p = 1/3.

Getting 'the multiple of 3' 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 = 180
p = 1/3

So write
B(180, 1/3).

B(180, 1/3)
p = 1/3

Then q = 1 - 1/3 = 2/3.

B(180, 1/6)
n = 180
p = 1/3

q = 2/3

Then the standard deviation σ(X) is
√180⋅[1/3]⋅[2/3].

180⋅[1/3] = 60

Cancel the denominator 3
and reduce 60 to, 60/3, 20.

20
= 4⋅5
= 2²⋅5

Prime Factorization

√2²⋅5⋅2
= 2√5⋅2
= 2√10

Simplify a Radical

So σ(X) = 2√10.