Probability 1 Question 9
9. 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
(2012)
(a) $\frac{91}{216}$
(b) $\frac{108}{216}$
(c) $\frac{125}{216}$
(d) $\frac{127}{216}$
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Answer:
Correct Answer: 9. (a)
Solution:
- PLAN As one of the dice shows a number appearing on one of $P _1, P _2$ and $P _3$.
Thus, the favourable number of cases $=9^{7}-8^{7}$.
$\therefore$ Required probability $=\frac{9^{7}-8^{7}}{15^{7}}$
