Question

Four fair dice $D_1$, $D_2$, $D_3$, $D_4$ each having faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1$, $D_2$, $D_3$ is
A) $\dfrac{{91}}{{216}}$
B) $\dfrac{{108}}{{216}}$
C) $\dfrac{{125}}{{216}}$
D) $\dfrac{{127}}{{216}}$

Answer 1

You’re adding up three probabilities of events that aren’t disjoint. D4 could show the same number as both D1 and D2, and you’re counting that possibility twice. Try using the inclusion-exclusion principle.
As one of the dice shows a number appearing on one of P1,P2 and P3
Thus, three cases arise.
All show the same number.
Number appearing on $D_4$ appears on any one of $D_1$, $D_2$ and $D_3$
Number appearing on $D_4$ appears on any two of $D_1$, $D_2$ and $D_3$

Complete step-by-step answer:
Sample space $= {\text{ }}6{\text{ }} \times {\text{ }}6{\text{ }} \times {\text{ }}6{\text{ }} \times 6{\text{ }} = {\text{ }}{{\text{6}}^4}$favourable events
= Case I or Case II or Case III
Case I First we should select one number for D4 which appears on all i.e. ${}^6{C_1}\; \times \;1$
Case II For D4 there are ${}^6{C_1}$ways. Now, it appears on any one of D1, D2 and D3 i.e., ${}^3{C_1}\; \times \;1$.
For the other two there are 5 x 5 ways. $= {}^6{C_1} \times {}^3{C_1}\; \times \;5\; \times \;5$
Case III For D4 there are ${}^6{C_1}$ways now it appears on any two of $D_1$, $D_2$ and $D_3$ $= {}^3{C_2}\; \times \;{1^2}$
For other one there are 5 ways $= {}^6{C_1} \times {}^3{C_2}\; \times \;{1^2}\; \times \;5$
Thus, probability $= \dfrac{{{}^6{C_1} + {}^6{C_1} \times {}^3{C_1}\; \times 5\; \times \;5 + {}^6{C_1} \times {}^3{C_2}\; \times \;5}}{{{6^4}}}$
$= \dfrac{{6 + 6 \times 3\; \times \;5\; \times \;5 + 6 \times 3\; \times \;5}}{{{6^4}}}$
$= \dfrac{{6 + 6 \times 75 + 6 \times 15}}{{{6^4}}} = \dfrac{{91}}{{216}}$

So, option (A) is the correct answer.

Note: Probability of occurrence of Event 'E'
$P(E) = \dfrac{{No.\;of\;favourable\;outcomes}}{{Total\;No.\;of\;outcomes}}$
Sum of Probabilities = 1, n(S) = 1
• Total No. of Outcomes when 'n' dice are rolled = $6^n$
• P(at least k times occurrence of Event E) = 1 - P(None of the times Event E occurs)