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. $\frac { 16 } { 81 }$
• B. $\frac { 1 } { 81 }$
• C. $\frac { 80 } { 81 }$
• D. $\frac { 65 } { 81 }$

Answer 1

Answer: (A)

$\frac { 16 } { 81 }$

Answer 2

P(minimum face value is not less than 2 and maximum face value is not greater than 5)

= P(2 or 3 or 4 or 5) $= \frac { 4 } { 6 } = \frac { 2 } { 3 }$ .

Hence required probability $= { } ^ { 4 } C _ { 4 } \left( \frac { 2 } { 3 } \right) ^ { 4 } \left( \frac { 1 } { 3 } \right) ^ { 0 } = \frac { 16 } { 81 }$.