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- 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 =\(1\over 3\)
