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/ n ). If the die is thrown thrice, then the probability, that the sum of the number obtained is 48 , is

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/ n ). If the die is thrown thrice, then the probability, that the sum of the number obtained is 48 , is (A) (7/211) (B) (7/212) (C) (3/210) (D) (13/212)

Tardigrade Admin · Accepted Answer

P(n)=(1/n) P(2)=(1/2) P(8)=(1/8) P(4)=(1/4) P(16)=(1/16) P(32)=(2/32) Possible cases 16,16,16 and 32,8,8 Probability =(1/163)+(2/32) × (1/8) × (1/8) × 3=(13/163)

Q. 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 number obtained is 48 , is

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

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

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

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

$P (n) = \frac{1}{n}$
$P (2) = \frac{1}{2} P (8) = \frac{1}{8}$
$P (4) = \frac{1}{4} P (16) = \frac{1}{16}$
$P (32) = \frac{2}{32}$
Possible cases
$16, 16, 16$ and $32, 8, 8$
Probability $= \frac{1}{1 6 ^{3}} + \frac{2}{32} \times \frac{1}{8} \times \frac{1}{8} \times 3 = \frac{13}{1 6 ^{3}}$

Q. 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 n1​. If the die is thrown thrice, then the probability, that the sum of the number obtained is 48 , is

2117​

2127​

2103​

21213​