A person draws a card from a pack of playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade. The chance that he will fail the first two times is

Options

A. `(1)/(16)`
B. `(9)/(16)`
C. `(9)/(64)`
D. `(1)/(64)`

Answer

B

Explanation

Let `A_(i)` be the event that the person fails in `i^(th)` draw.
Then,
`P(A_(i))=1-(13)/(52)=(3)/(4), i=1,2,3,..`
`therefore` Required Probability `=P(A_(1) cap A_(2))=P(A_(1))P(A_(2))`
`implies` Required probability `=(3)/(4)xx(3)/(4)=(9)/(16)`