Three dice are thrown. The probability that the sum of the numbers appearing is 15, is :

Options

A. `(7)/(216)`
B. `(47)/(54)`
C. `(7)/(54)`
D. `(7)/(9)`

Answer

C

Explanation

Total number of elementary events `=6^(3)`.
Total number of ways of getting 9 as the sum
= Coefficient of `x^(9) " in " (x^(1)+x^(2)+x^(3)+x^(4)+x^(5)+x^(6))^(3)`
= Coefficient of `x^(6) " in " (1+x+x^(2)+x^(3)+x^(4)+x^(5))^(3)`
= Coefficient of `x^(6) " in " ((1=x^(6))/(1-x))^(3)`
= Coefficient of `x^(6) " in " (1-x)^(-3)= .^(6+3-1)C_(3-1)= .^(8)C_(2)=28`
`therefore` Favourable number of elementary events =28.
So, required probability `=(28)/(216)=(7)/(54)`