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 :
Solution
<p>There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)</p>
<p>So, required probability</p>
<p>$$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$$</p>
<p>$= {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$</p>
<p>$= {{13} \over {{2^{12}}}}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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