A biased coin with probability `p(0 lt plt 1)` of falling tails is tossed until a tail appears for the first time. If the probability that tail comes in odd number of trials is `(2)/(3)`, then `p` equals

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(3)/(4)`
D. `(1)/(2)`

Answer

D

Explanation

`(d)` `P(T)=p`, `P(H)=1-p`
`:.` Required probability
`=P(T or HHT or HHHHT or ….)`
`=p+(1-p)^(2)p+(1-p)^(4)+….`
`=(p)/(1-(1-p)^(2))`
`=(p)/(2p-p^(2))=(2)/(3)`
`impliesp=(1)/(2)`