Question

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

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

Answer 1

Answer: (A)

Since, $n(S)=6^4=1296$
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
$=\frac{12+24+12+12+24+12+12+12+4+4+12}{1296}$
$=\frac{140}{1296}=\frac{35}{324}$