Question

If four dice are thrown together, then the probability that the sum of the numbers appearing on them is 13 is

• A. $\frac{5}{216}$
• B. $\frac{35}{324}$
• C. $\frac{11}{432}$
• D. $\frac{11}{216}$

Answer 1

The correct option is B $\frac{35}{324}$

since $n\left(S\right)=6^4=1296$ and permutation that the sum of the numbers appearing on them is $13$.
Total permutation of $(1,1,5,6)=\frac{4!}{2!}=12$
Total permutation of $(1,2,4,6)=4!=24$
Similarly for $(1,3,3,6)=\frac{4!}{2!}=12$
$(1,2,5,5)=12;(1,3,5,4)=24;(2,2,6,3)=12;(2,2,5,4)=12$;
$(3,3,2,5)=12;(3,3,3,4)=4;(4,4,4,1)=4$and$4,4,3,2)=12$
$\therefore$ Required probability
$=\frac{12+24+12+12+24+12+12+12+4+4+12}{1296}=\frac{140}{1296}=\frac{35}{324}$