Three dice are thrown together. The probability that the sum of the numbers appearing on them is 9, is
Answer & explanation
Correct answer: option 3
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 $\left(\frac{1-x^6}{1-x}\right)^3$
= Coefficient of $x^6$ in $(1-x)^{-3}= {^{6+3-1}C}_{3-1}= {^8C}_2= 28 $
∴ Favourable number of elementary events = 28.
So, required probability $=\frac{28}{216}=\frac{7}{54}$