Mark the correct alternative in the following:

A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p equals.

Given:

Probability of getting head to appear first time(r=1), n should be even.

So we have to check probability for n=2,4,6…

P(x=r)= ⁿC_r p^r q^n-r = 2/5

For n=2;

P(X=1)= pq

For n=4

P(X=1) = pq³

Required probability = pq + pq³ + pq⁵ + … =2/5

2/5= pq(1+ q² +q⁴ + …)

4-2p=5-5p

3p= 1

P =1/3