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Question
An unbiased die with face marked 1, 2, 3, 4, 5, 6 is rolled four times. Out of 4 face values obtained, find the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5.
Solution
\[P\left( \text{ face value is not more than 5 and not less than } 2 \right) = \frac{4}{6} = \frac{2}{3}\]
\[P\left( \text{ face value is not more than 5 and not less than 2 in 4 throws } \right) = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\]
\[ = \frac{16}{81}\]
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