A biased coin with probability p, `0ltplt1` 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

Explanation

Chance of doing 2 toss `=(1-p)^{*} p`

Chance of doing 4 toss `=(1-p)^{*}(1-p)^{*}(1-p)^{*} p`

Chance of doing 6 toss `=(1-p)^{*}(1-p)^{*}(1-p)^{*}(1-p)^{*}(1-p)^{*} p`

Probability that number of coins is even `=` chance of `2+` chance of `4+` chance of `6+ldots ldots`

`=(1-p) * p(1+(1-p) *(1-p)+(1-p) *(1-p) *(1-p) *(1-p)+ldots .)`

`Rightarrow(1-p) * p(frac{1}{1-(1-p)^{2}})=2 / 5`

`Rightarrow(1-p) * p(frac{1}{p(2-p)})=2 / 5`

`Rightarrow(1-p) *(frac{1}{(2-p)})=2 / 5`

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