Three-six-faced-dice-are-thrown-together-The-probability-that-the-sum-of-the-numbers-appearing-on-the-dice-is-k-3-k-8-is-

Question Number 47481 by limite last updated on 10/Nov/18

Three six - faced dice are thrown

together . The probability that the sum

of the numbers appearing on the dice

is k (3 ⩽ k ⩽ 8) is

Answered by ajfour last updated on 10/Nov/18

c o e f f . o f x^{3} + c o e f f . o f x^{4} + \dots .

\dots + c o e f f . o f x^{8} i n t h e e x p a n s i o n

o f {(x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})}^{3} b e N

P (k) = \frac{N}{216} .

{(x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})}^{3}

= x^{3} \frac{{(1 - x^{6})}^{3}}{{(1 - x)}^{3}}

= x^{3} (1 - 3 x^{6} + \dots) {(1 - x)}^{- 3}

N = c o e f f . o f x^{k} w h e r e 3 ⩽ k ⩽ 8 i s

= c o e f f . o f x^{k - 3} i n t h e e x p a n s i o n

o f {(1 - x)}^{- 3}

= Σ^{3 + k - 3 - 1} C_{3 - 1} = Σ^{k - 1} C_{2}

=^{7} C_{2} +^{6} C_{2} +^{5} C_{2} +^{4} C_{2} +^{3} C_{2} +^{2} C_{2}

= 21 + 15 + 10 + 6 + 3 + 1 = 56

P (k) = \frac{56}{216} = \frac{7}{27} .

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