Question

Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is K $(3\le K\le 8)$ is

• A. $\frac{(k-1)(k-2)}{432}$
• B. $\frac{k(k-1)}{432}$
• C. $\frac{k^2}{432}$
• D. None of these

Answer 1

Answer: (A)

$\frac{(k-1)(k-2)}{432}$

$given3\le K\le 8K=3canbeonlywhen1appearsoneachdieP(K=3)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}P(K=4)=\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{3}{216}\{permutationof1,1,2onthedice\}P(K=5)=\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{6}{216}\{permutationof3,1,1and2,2,1onthedice\}P(K=6)=\left(\begin{array}{c}3\\ 1\end{array}\right)\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+3!\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{10}{216}\{permutationof1,1,41,2,3and2,2,2onthedice\}observingthepatterninnumeratorofP\left(K\right)wecanwriteP(K=7)=\frac{15}{216}P(K=8)=\frac{21}{216}ThenumeratorofP\left(K\right)is1,3,6,10,15,21sowecanwritegeneraltermforthiswhen3\le K\le 8LetS=1+3+6+........+T_{n-1}+T_n\Rightarrow S=1+3+.........+T_{n-2}+T_{n-1}+T_{n}subtractingtheabovetwoequationsandrearrangingweget{T}_n=1+2+3+......+n\Rightarrow T_n=\frac{n\times (n+1)}{2}nowputtingK=n+2\Rightarrow n=K-2\because Kstartswith3T_k=\frac{(K-2)\times (K-1)}{2}whichisthegeneraltermfornumeratorofP\left(K\right)\therefore P\left(K\right)=\frac{T_k}{216}=\frac{(K-2)\times (K-1)}{432}$