Question

Six faces of a die are marked with the numbers

The die is thrown thrice. The probability that the sum of the numbers thrown is six, is

• A. $\frac { 1 } { 72 }$
• B. $\frac { 1 } { 12 }$
• C. $\frac { 5 } { 108 }$
• D. $\frac { 1 } { 36 }$

Answer 1

Answer: (C)

$\frac { 5 } { 108 }$

Answer 2

Total number of elementary events $= 6 ^ { 3 }$

Favourable number of elementary events.

$=$ Coeff. of $x ^ { 6 }$ in $\left( x + x ^ { - 1 } + x ^ { 0 } + x ^ { - 2 } + x ^ { 2 } + x ^ { 3 } \right) ^ { 3 }$

$=$ Coeff.of $x ^ { 12 }$ in $\left( 1 - x ^ { 6 } \right) ^ { 3 } ( 1 - x ) ^ { - 3 }$

$= { } ^ { 12 + 3 - 1 } \mathrm { C } _ { 3 - 1 } - { } ^ { 3 } \mathrm { C } _ { 1 } \cdot { } ^ { 6 + 3 - 1 } \mathrm { C } _ { 3 - 1 } + { } ^ { 3 } \mathrm { C } _ { 2 }$

$= { } ^ { 14 } \mathrm { C } _ { 2 } - { } ^ { 3 } \mathrm { C } _ { 1 } \times { } ^ { 8 } \mathrm { C } _ { 2 } + { } ^ { 3 } \mathrm { C } _ { 2 } = 91 - 84 + 3 = 10$

Hence, required probability $= \frac { 10 } { 6 ^ { 3 } } = \frac { 5 } { 108 }$