Probability 3 Question 17
17. An unbiased die with faces marked $1,2,3,4,5$ and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 , is
(a) $16 / 81$
(b) $1 / 81$
(c) $80 / 81$
(d) $65 / 81$
(1993, 1M)
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Solution:
- Let $A=$ getting not less than 2 and not greater than 5
$$ \Rightarrow \quad A={2,3,4,5} \quad \Rightarrow P(A)=\frac{4}{6} $$
But die is rolled four times, therefore the probability in getting four throws
$$ =\begin{array}{llll} \frac{4}{6} & \frac{4}{6} & \frac{4}{6} & \frac{4}{6} \end{array}=\frac{16}{81} $$
