Question

Four fair dice $D_1,D_2,D_{3}and{D}_4$ each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1,D_{2}and{D}_3$, is ?

• A. $\frac{91}{216}$
• B. $\frac{108}{216}$
• C. $\frac{125}{216}$
• D. $\frac{127}{216}$

Answer 1

The correct option is A

$\frac{91}{216}$

Sample space $6\times 6\times 6\times 6=6^4$ favourable events
= Case I or Case II or Case III
Case I First we should select one number for $D_4$ which apperars on all i.e. ${}^6{C}_1\times 1.$
Case II For $D_4$ there are ${}^6{C}_1$ ways. Now, it appears on any one of $D_1,D_2$ and $D_3$ i.e. ${}^3{C}_1\times 1$.
For other two there are $5\times 5$ ways.
${\Rightarrow }^6{C}_1{\times }^3{C}_1\times 1\times 5\times 5$
Case III For $D_4$ there are ${}^6{C}_1$ ways now it appears on any two of $D_1,D_{2}and{D}_3$
${\Rightarrow }^3{C}_2\times 1^2$
For other one there are 5 ways.
${\Rightarrow }^6{C}_1{\times }^3{C}_2\times 1^2\times 5$
Thus, probability $=\frac{{}^6{C}_1{+}^6{C}_1{\times }^3{C}_1\times 5^2{+}^6{C}_1{\times }^3{C}_2\times 5}{6^4}$
$\frac{6(1+75+15)}{6^4}=\frac{91}{216}$