Question

An unbaised die with faces marked $1,2,3,4,5$ and $6$ is rolled four times out of the four faces obtained, the probability that the minimum value is exactly than 2 and the maximum value is exactly $5$ is

• A. $\frac{4}{27}$
• B. $\frac{1}{81}$
• C. $\frac{80}{81}$
• D. $\frac{65}{81}$

Answer 1

The correct option is A $\frac{4}{27}$

Total number of outcomes of 4 dice is $6^4$. We can choose two dice
(one for 2 and one for 5) in ${}^4{C}_2$ ways. We can arrange 2 and 5 at
these places in 2 ways. For the remaining two dice, the number of
possible outcomes is $\left(4\right)\left(4\right)=16$. Thus, the number of favourable
outcomes is ${(}^4{C}_2\left)\right(2\left)\right(16)$. Thus, probability of required event
$=\frac{\left(6\right)\left(2\right)\left(16\right)}{6^4}=\frac{4}{27}$.