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 numbers obtained is 48, is

Correct answer: (D) 132^12. P2=1/2,P4=1/4,P8=1/8,P16=1/16,P32=1/32 When we get the faces as 8,8,32 then the probability will be 31/8×1/8×2/32=3/1024 When we get the faces as 16,16,16 then the probability will be 1/16×1/16×1/16=1/4096 Hence, the required probability will be3/1024+1/4096=13/4096

Q45 / 90 · JEE Main 2022 (25 Jun Shift 2) · Mathematics

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
$\frac{3}{2^{10}}$
Correct answer
D
$\frac{13}{2^{12}}$

Solution

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

When we get the faces as $8, 8, 32$ then the probability will be $3 (\frac{1}{8} \times \frac{1}{8} \times \frac{2}{32}) = \frac{3}{1024}$

When we get the faces as $16, 16, 16$ then the probability will be $\frac{1}{16} \times \frac{1}{16} \times \frac{1}{16} = \frac{1}{4096}$

Hence, the required probability will be $\frac{3}{1024} + \frac{1}{4096} = \frac{13}{4096}$

Hence, the correct answer is option: 'D'.

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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 1n. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is

Solution

Relevant Questions

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