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

Answer: (A)

PLAN $ $ As one of the dice shows a number appearing on one of $P_1,P_2$ and $P_3$
Thus, three cases arise.
(i) All show same number.
(ii) Number appearing on $D_4$ appears on any one of
$D_1,D_2 \,$ and $\, D_3$
(iii) Number appearing on $D_4$ appears on any
two of $D_1, D_2$ and $D_3$
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 appears on all i.e. ${}^6 C_1 \times 1$
Case II For $D_4$ there are ${}^6C_1$ ways. Now, it appears
on any one of $D_1 ,D_2 \,$ and $\, D_3 \, i.e., \, \,{}^3C_1 \times \, 1.$
For other two there are $5 \times 5$ ways.
$\Rightarrow {}^6C_1 \times \,{}^3C_1 \, \times \, 5 \, \times 5$
Case III For $D_4$ there are $^6C_1$ 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
$\Rightarrow {}^6C_1 \, \times \,{}^3C_2 \, \times \, 1^2 \, \times 5$
Thus, probability =$\frac{\,{}^6C_1+ \,{}^6C_1 \times \,{}^3C_1 \times 5^2 \,{}^6C_1 \times \,{}^3C_2 \times 5}{6^4}$
$ =\frac{6(1+75+15}{6^4}=\frac{91}{216}$