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

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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- p

\(\implies\) \(2\over 5\) = \((1 – p)p\over {1 – (1 – p)^2}\)

\(\implies\) \(2\over 5\) = \((1 – p)\over {2 – p}\)

\(\implies\) 4 – 2p = 5 – 5p

\(\implies\) p =\(1\over 3\)

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