A biased coin with probability `p, 0 lt p lt 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

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 i.e., 2 or 4 or 6 etc.
`therefore (2)/(5)=ap+q^3p+q^5p+......`
`rArr (2)/(5)=(qp)/(1-q^2)`
`rArr (2)/(5)=((1-p)p)/(1-(1-p)^2)=(1-p)/(2-p)`
`rArr 2(2-p)=5(1-p)rArr p=1//3`