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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 \mathrm{n} is \mathrm{\frac{1}{n}}. If the die is thrown thrice, the the probability,that the sum of the numbers obtained is \mathrm{48} is :

Option: 1

\frac{7}{2^{11}}


Option: 2

\frac{7}{2^{12}}


Option: 3

\frac{3}{2^{10}}


Option: 4

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


Answers (1)

best_answer

\mathrm{ 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 \mathrm{8,8,32} the the probability will be \mathrm{3\left ( \frac{1}{8}\times\frac{1}{8}\times\frac{2}{32} \right )= \frac{3}{1024}}

Whe we get the faces as\mathrm{16,16,16} then the proability will be \mathrm{\frac{1}{16}\times\frac{1}{16}\times\frac{1}{16}= \frac{1}{4096}}

Hece, the required proaility will be

\mathrm{\frac{3}{1024}+\frac{1}{4096}=\frac{13}{4096}= \frac{13}{2^{12}}}

 Hence the correct answer is option 4

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