When a coin is tossed, the probability of getting a Head is p,0p1. 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/p2
(D) 1/(1−p2)
Answer: (A)
Explanation:
For a continuous variable X ranging over all the real numbers, the expectation is defined by
E(X)= ∫ xf(x) dx
probability for head = p
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
correct answer is A.
This solution is contributed by Nitika Bansal.
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