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

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(2)/(5)`
D. `(3)/(5)`

Answer

A

Explanation

Let `q=1-p`. Since, head appears first time in an even throw 2 or 4 or 6 and so on.
`therefore (2)/(5)=qp + q^(3)p+q^(5)p+ … rArr (2)/(5)=(qp)/(1-q^(2))`
`rArr (2)/(5)=((1-p)p)/(1-(1-p)^(2))rArr (2)/(5)=(1-p)/(2-p)`
`rArr 4-2p=5-5p rArr p=(1)/(3)`