A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability o…

Question

A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is $${1 \over n}$$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

Accepted Answer

There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)

So, required probability

$$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$$

$$ = {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$$

$$ = {{13} \over {{2^{12}}}}$$

Solutions · 1