Answer:
Option A
Explanation:
Concept involved As one of the dices shows a number appearing on one of $P_{1},P_{2} $ and $P_{3}$
Thus, three cases arise
(i) All shows the same number
(ii) Number appearing on $D_{4}$ appears on any one of $D_{1},D_{2}, D_{3}$
(iii) Number appearing on $D_{4}$ appears on any two of $D_{1},D_{2},D_{3}$
sol .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 $^{6}C_{1}$ ways .Now, it appears on any one of $D_{1},D_{2},D_{3}$
i.e, $^{3}C_{1} \times 1$ for other two there are 5 x 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 $D_{1},D_{2},D_{3} \Rightarrow ^{3}C_{2} \times 1^{2}$
for other one there are 5 ways
$\Rightarrow$ $^{6}C_{1} \times ^{3}C_{2} \times 1^{2} \times 5 $
Thus probability
$=\frac{^{6}C_{1}+^{6}C_{1} \times^{3}C_{1}\times 5^{2}+^{6}C_{1}\times ^{3}C_{2}\times 5}{6^{4}}$
$=\frac{6(1+75+15)}{6^{4}}=\frac{91}{216}$
