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 $\frac1{n}$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48 is:

• A. $\frac{7}{2^{11}}$
• B. $\frac{7}{2^{12}}$
• C. $\frac{3}{2^{10}}$
• D. $\frac{13}{2^{12}}$

Answer 1

Answer: (D)

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

Answer 2

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

So, the required probability

$=3(\frac{2}{32}.\frac{1}{8}.\frac{1}{8})+(\frac{1}{16}.\frac{1}{16}\frac{1}{16})$

$=3(\frac{1}{32}\,. \,\frac{1}{32})+\frac{1}{16^3}=\frac{3}{2^{10}}+\frac{1}{2^{12}}$

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

Therefore, The correct option is(D) : $\frac{13}{2^{12}}$