GATE | GATE IT 2006 | Question 22

• Last Updated : 02 Dec, 2021

When a coin is tossed, the probability of getting a Head is . Let 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 is
(A) 1/p
(B) 1/(1−p)
(C) 1/p2
(D) 1/(1−p2)

Explanation:

For a continuous variable X ranging over all the real numbers, the expectation is defined by

E(X)= ∫ xf(x) dx

probability for tail = 1-p
if in first time, head appears, probability will be 1*p
if firstly tail occurs,and then head occurs, then the probability will be (1-p)*p
and so on…. for the nth time, probability will be (1-p) n-1 * p
E= 1*p + 2*(1−p)*p + 3*(1−p)*(1−p)*p + ………………. equation(1)
multiply both side with (1−p):
E*(1-p) = 1*p*(1-p) + 2*(1-p)*(1-p)*p + 3*(1-p)*(1-p)*(1-p)*p +…………. equation (2)
Subtract equation 2 from equation 1:

E−E*(1−p)= 1*p+ (1−p)*p+ (1−p)*(1−p)*p +…
E*p =p[1+ (1-p) + (1-p)*(1-p) + ……]

It’s a infinite geometric progression.
E = 1/(1-(1-p)) = 1/p
E=1/p