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Question: Three six faced fair dice are thrown together. The probability that the sum of the numbers appearing...

Three six faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(3≤k≤8)k ( 3 \leq k \leq 8 ) , is

A

(k−1)(k−2)432\frac { ( k - 1 ) ( k - 2 ) } { 432 }

B

k(k−1)432\frac { k ( k - 1 ) } { 432 }

C

k2432\frac { k ^ { 2 } } { 432 }

D

None of these

Answer

(k−1)(k−2)432\frac { ( k - 1 ) ( k - 2 ) } { 432 }

Explanation

Solution

The total number of cases =6×6×6=216= 6 \times 6 \times 6 = 216 The number of

favourable ways

= Coefficient of xkx ^ { k } in = Coefficient of

xk−3x ^ { k - 3 } in (1−x6)3(1−x)−3\left( 1 - x ^ { 6 } \right) ^ { 3 } ( 1 - x ) ^ { - 3 } = Coefficient of xk−3x ^ { k - 3 } in (1−x)−3( 1 - x ) ^ { - 3 }

{0≤k−3≤5}\{ 0 \leq k - 3 \leq 5 \}

=Coefficient of xk−3x ^ { k - 3 } in

(1+3C1x+4C2x2+5C3x3+….)=k−1C2=(k−1)(k−2)2\left( 1 + { } ^ { 3 } C _ { 1 } x + { } ^ { 4 } C _ { 2 } x ^ { 2 } + { } ^ { 5 } C _ { 3 } x ^ { 3 } + \ldots .\right) = { } ^ { k - 1 } C _ { 2 } = \frac { ( k - 1 ) ( k - 2 ) } { 2 }

Thus the probability of the required event is (k−1)(k−2)432\frac { ( k - 1 ) ( k - 2 ) } { 432 } .