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 $\frac{1}{n}$ . If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48 , is

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$\frac{3}{2^{10}}$

$\frac{7}{2^{12}}$

$\frac{13}{2^{12}}$

$\frac{7}{2^{11}}$

answer is D.

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Detailed Solution

$P (n) = \frac{1}{n} P (2) = \frac{1}{2} P (8) = \frac{1}{8} P (4) = \frac{1}{4} P (16) = \frac{1}{16} P (32) = \frac{2}{32}$

Possible cases

16,16,16 and 32,8,8

Probability = $\frac{1}{16^{3}} + \frac{2}{32} \times \frac{1}{8} \times \frac{1}{8} \times 3 = \frac{13}{16^{3}}$

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