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, the probability, that the sum of the numbers obtained is 48, is m then the nearest integral value of 210m is

Possible cases 16,16,16and32,8,8Probability =1163+232×18×18×3=13163m=13212⇒210m=134

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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 $\frac{1}{n}$ . If the die is thrown thrice, the probability, that the sum of the numbers obtained is $48$ , is $m$ then the nearest integral value of $2^{10} m$ is

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Possible cases $16, 16, 16 a n d 32, 8, 8$

Probability $= \frac{1}{16^{3}} + \frac{2}{32} \times \frac{1}{8} \times \frac{1}{8} \times 3 = \frac{13}{16^{3}}$

$m = \frac{13}{2^{12}} \Rightarrow 2^{10} m = \frac{13}{4}$

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