A biased die is marked with numbers 2, 4, 8, | vedclass

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(D) The probabilities of getting each face are given by $P(n) = \frac{1}{n}$.$P(2) = \frac{1}{2}, P(4) = \frac{1}{4}, P(8) = \frac{1}{8}, P(16) = \fra

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

Vedclass · Answer

(D) The probabilities of getting each face are given by $P(n) = \frac{1}{n}$.$P(2) = \frac{1}{2}, P(4) = \frac{1}{4}, P(8) = \frac{1}{8}, P(16) = \frac{1}{16}, P(32) = \frac{2}{32} = \frac{1}{16}$.We need the sum of three throws to be $48$. The possible combinations are:$1$. $(16, 16, 16)$: Probability $= P(16) 	imes P(16) 	imes P(16) = \frac{1}{16} 	imes \frac{1}{16} 	imes \frac{1}{16} = \frac{1}{16^3} = \frac{1}{4096}$.$2$. $(32, 8, 8)$ in any order: The number of permutations is $\frac{3!}{2!} = 3$.Probability $= 3 	imes P(32) 	imes P(8) 	imes P(8) = 3 	imes \frac{1}{16} 	imes \frac{1}{8} 	imes \frac{1}{8} = \frac{3}{1024} = \frac{12}{4096}$.Total probability $= \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096} = \frac{13}{2^{12}}$.

$A$ biased die is marked with numbers $2, 4, 8, 16, 32, 32$ on its faces. 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:

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