Two persons `A` and `B` get together once a weak to play a game. They always play `4` games . From past experience Mr. A wins `2` of the `4` games just as often as he wins `3` of the `4` games. If Mr. A does not always wins or always loose, then the probability that Mr. A wins any one game is (Given the probability of A's wining a game is a non-zero constant less than one).

Options

A. `0.5`
B. `0.6`
C. `0.8`
D. `0.9`

Answer

B

Explanation

`(b)` `"^(4)C_(2)p^(2)(1-p)^(2)=^(4)C_(3)p^(3)(1-p)`
`implies6(1-p)=4p`
`impliesp=(3)/(5)=0.6`