JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

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 $\dfrac1n$ . If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

\dfrac7{2^{11}}

\dfrac7{2^{12}}

\dfrac3{2^{10}}

\dfrac{13}{2^{12}}

Answer and worked solution

Official answer: D

Face probabilities. $P(2)=\frac12,P(4)=\frac14,P(8)=\frac18,P(16)=\frac1{16}$ , and 32 (two faces) $\frac1{16}$ .

Sum 48 from three throws: $16+16+16$ or $32+8+8$ (in 3 orders).

Probability. $\left(\frac1{16}\right)^3+3\cdot\frac1{16}\cdot\frac1{64}=\frac1{4096}+\frac{12}{4096}=\frac{13}{2^{12}}$ .

(D)

Topics:Classical probability Independent events

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 16