Question

A fair die is rolled until the number 6 first appears. Let N be the number of rolls, including the last roll. Given that we have rolled at least 4 times, what is the expected number of rolls?

This one's killing me

Answer 1

Ans:

Probability (p) of getting a 6 in throwing a die is 1/6  and the probability (q) of not getting a 6 is 1−1/6=5 /6.

Let Nth throw of the die results in a 6. This means that preceding (N -1) throws do not result in a 6.

So, the probability function is p(N)=qN−1 p for N= 0,1,2,…….

Hence, E(N)=∑∞N-1 N q N−1 p

=p∑∞N-1 N qN−1  

=p(1+2q+3q2+4q3+....)=p(1+2q+3q2+4q3+....)

=p(1−q)−2=p(1−q)−2

=16(1−56)−2=16(1−56)−2

=(16)(16)−2=(16)(16)−2

=16∗36=6

expected numbers of rolls =6