An unbiased dice with faces marked 1, 2,3, 4,5 , and 6 is rolled four times. Out of four face values…
An unbiased dice 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
$16 / 81$
$1 / 81$
$80 / 81$
$65 / 81$
Solution
Here, the face value should be 2, 3, 4, 5 . $P$ (getting a number not less than 2 and not more than 5 in a single throw) $=\frac{4}{6}=\frac{2}{3}$ $\therefore$ Required probability $=\left(\frac{2}{3}\right)^4=\frac{16}{81}$