Question

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. Jan-81
• C. 80/81
• D. 65/81

Answer 1

Answer: (A)

16/81

Answer 2

Let A = getting not less than 2 and not greater than 5
$\rightarrow \, \, \, \, \, \, \, \, \, \, \, A=\\{2,3,4,5 \\} \, \, \Rightarrow \, P(A)=\frac{4}{6}$
But die is rolled four times, therefore the probability in getting four throws
$\, \, \, \, \, \, \, \, \, \, \, =\bigg(\frac{4}{6}\bigg)\bigg(\frac{4}{6}\bigg)\bigg(\frac{4}{6}\bigg)\bigg(\frac{4}{6}\bigg) =\frac{16}{81}$