First slide

Introduction to probability

Question

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

Moderate

A

$\frac{1}{6}$
B

$\frac{1}{36}$
C

$\frac{12}{151}$
D

none of these

Solution

Given that 5 and 6 have appeared on two of the dice,
the sample space reduces to 6⁴ - 2 x 5⁴ + 4⁴ (inclusion-exclusion principle).
Also, the number of favorable cases are 4! = 24.
So, the required probability is $\frac{24}{302} = \frac{12}{151}$

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