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 a) 1/6 b) 1/36 c) 12/151 d) none of these

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`.