## 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 is equal to

Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. \(\therefore\) \(2\over 5\) = qp + \(q^3\)p + \(q^5\)p + …… \(\implies\) \(2\over 5\) = \(qp\over {1 – q^2}\) Since q = 1- …