Question

Find the probability of getting at least twenty-five 4's in 90 rolls of a 6-sided die. Justify any approximation that you use.

Answer 1

p = P(Getting a 4 in single throw) = 1/6

n = Number of times die are rolled = 90

Let the random variable X is defined as

X : Number of times getting 4 in 90 rolls.

So random variable X takes values 0,1,2,.........,90.

Since p is constant for each roll and each throw are independent to each other.

Hence probability distribution of X is binomial with parameters n = 90 and p = 1/6

X ~ Bin ( n=90, p = 1/6)

E(X) = n* p*= 90 * (1/6)  = 15

var(X) = n*p * (1-p) = 90 * (1/6) * ( 5/6) =12.5

SD(X)= sqrt(12.5) =3.5355.

P ( at least twenty-five 4's in 90 rolls) = P ( X> =25)

Since n is large , we use normal approximation to binomial distribution .

By continuity correction

= P ( Z> 2.6870)

From normal probability table

P( Z> 2.6870) =0.0036

P ( at least twenty-five 4's in 90 rolls) = 0.0036.