Question

a) With a fixed probability of p for scoring three pointers, a basketball player takes 5 independent shots. What is the probability that he scores 1 shot in the first 3, and 1 again in the last 2 shots? b) In a series of indepedent and identically distributed Bernoulli trials, why is the index of the second successful trial not a geometric random variable? Explain.

Answer 1

Answer a:

Probability of success = p

Probability of failure = 1 - p

Probability that he scores 1 shot in the first 3 = (3C1) x p x (1 - p)^2

(Since, he can have success in any shot, can be the first or the second or the third)

Also, probability that he scores 1 shot in the next 2 = (2C1) x p x (1 - p)

Total required probability = 6(p^2) ((1 - p) ^3)

Answer b:

Let probability of success be p

Therefore, probability of failure = (1 - p)

Suppose,

Let the second success come in the kth trial

Which means in the previous (k - 1) trials there was 1 success and (k - 2) number of failures.

Number of ways of attaining 1 success in (k - 1) trials = (k - 1)C1 = (k - 1)

Therefore,

Probability of second success in the kth trial = (k - 1)p x (1-p)^(k-2) x p
which does not follow geometric distribution.

Thus, the index of a second successful trial is not a geometric random variable