Four die are thrown simultaneously. The probability that `4` and `3` appear on two of the die given that `5` and `6` have appeared on other two die is

Options

A. `1/6`
B. `1/36`
C. `12/151`
D. none of these

Answer

C

Explanation

Given that 5 and 6 have appeared on two of the dice, the sample space reduces to `6^(4) - 2 xx 5^(4) + 4^(4)` (inclusion-exclusion principle). Also, the number of favorable cases are `4! = 24`. So, the required probability is `24//302 = 12//151`.