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

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Probability

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

\frac{1}{n}

. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48 , is

Answer: (D)

The probabilities are:

P (2) = \frac{1}{2}

P (4) = \frac{1}{4}

P (8) = \frac{1}{8}

P (16) = \frac{1}{16}

P (32) = \frac{2}{32} = \frac{1}{16}

(since 32 appears twice). To get sum 48 in three throws, the possible cases are: (16,16,16) and (32,8,8). For (16,16,16): probability

= (\frac{1}{16})^{3} = \frac{1}{4096}

. For (32,8,8): probability

= \frac{1}{16} \cdot \frac{1}{8} \cdot \frac{1}{8} \times 3 = \frac{3}{1024}

(the factor 3 accounts for the three orderings). Total probability

= \frac{1}{4096} + \frac{3}{1024} = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096} = \frac{13}{2 ^{12}}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option