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 numbers obtained is 48, is

(A) 7/2¹¹

(B) 7/2¹²

(C) 3/2¹⁰

(D) 13/2¹²

Answer 1

Correct option is (D) 13/2¹²

Probability = \(\frac1{16^3}+\frac2{32}\times\frac18\times18\times3\)

 = \(\frac {13}{16^3}\)