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:

  1. 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 $=\frac{9^{7}-8^{7}}{15^{7}}$



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