Question

21) Approximate the probability that in 205 tosses of a fair die, we will obtain at least 109 fives.

Answer 1

Solution:

Here, we have to use the normal approximation for finding required probability.

We have to find P(X≥109) = P(X>108.5) (by using continuity correction)

We are given

n = 205

p = P(5) = 1/6 = 0.1667

q = 1 - p = 1 - 0.1667 = 0.8333

Mean = np = 205*0.1667 = 34.1735

SD = sqrt(npq) = sqrt(205*0.1667*0.8333) = 5.336364

P(X>108.5) = 1 - P(X<108.5)

Z = (X - mean)/SD

Z = (108.5 - 34.1735)/ 5.336364

Z = 13.9283

P(Z<13.9283) = P(X<108.5) = 1

(by using z-table or excel)

P(X>108.5) = 1 - P(X<108.5)

P(X>108.5) = 1 - 1

P(X>108.5) = 0.0000

Required probability = 0.0000