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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 :

${7 \over {{2^{11}}}}$
${7 \over {{2^{12}}}}$
${3 \over {{2^{10}}}}$
${{13} \over {{2^{12}}}}$

Solution

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