When a coin is tossed, the probability of getting a Head is p (0 < p < 1). Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is: A. 1/p B. 1/(1-p) C. 1/p^2 D. 1/(1-p)^2

GATE 2006 · Engineering Mathematics · Binomial Distribution · medium

Answer: E[N] = 1/p. Answer: A.

1. Set up the expectation equation: Let E = E[N]. Condition on the first toss: P(Head) = p, then N = 1. P(Tail) = 1-p, then we have used 1 toss and restart with the same expected remaining E. So E = p * 1 + (1-p) * (1 + E).
2. Solve for E: E = p + (1-p) + (1-p)E = 1 + (1-p)E. Therefore E - (1-p)E = 1, so pE = 1, so E = 1/p.