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 $\frac{1}{n}$. 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)

$P ( n )=\frac{1}{ n }$
$P (2)=\frac{1}{2}\,\, P (8)=\frac{1}{8}$
$P (4)=\frac{1}{4}\,\, P (16)=\frac{1}{16}$
$P (32)=\frac{2}{32}$
Possible cases
$16,16,16$ and $32,8,8$
Probability $=\frac{1}{16^{3}}+\frac{2}{32} \times \frac{1}{8} \times \frac{1}{8} \times 3=\frac{13}{16^{3}}$