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)= nCr pr qn-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

1