If four dice are thrown together. Probability that the sum of the number appearing on them is 13, is

Options

A. `(35)/(324)`
B. `(5)/(216)`
C. `(11)/(216)`
D. `(11)/(432)`

Answer

A

Explanation

The total number of elementary events associated with the experiment of throwing four dice is `6xx6xx6xx6=6^(4)`.
Favourable number of elementary events
= Coefficient of `x^(13) " in " (x^(1)+x^(2)+x^(3)+x^(4)+x^(5)+x^(6))^(4)`
= Coefficient of `x^(9) " in " (1+x+x^(2)+..+ x^(5))^(4)`
= Coefficient of `x^(9) " in " ((1-x^(6))/(1-x))^(4)`
= Coefficient of `x^(9) " in " (1-x^(6))^(4) (1-x)^(-4)`
=Coefficient of `x^(9) " in " (1- .^(4)C_(1) x^(6)+ .^(4)C_(2)x^(12))(1-x)^(-4)`
= Coefficient of `x^(9) " in " (1-x)^(-4)- .^(4)C_(1)xx " Coefficient of " x^(3) " in " (1-x)^(-4)`
`= .^(9+4-1)C_(4-1)- .^(4)C_(1)xx .^(3+4-1)C_(4-1)`
`[because " Coeff. of " x^(n) " in" (1-x)^(-r)= .^(n+r-1)C_(r-1)]`
`= .^(12)C_(3)-4 xx .^(6)C_(3)=140`
`therefore` Required probability `=(140)/(6^(4))=(35)/(324)`