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 then:
A.
$$\frac{{16}}{{81}}$$
B.
$$\frac{{1}}{{81}}$$
C.
$$\frac{{80}}{{81}}$$
D.
$$\frac{{65}}{{81}}$$
Answer :
$$\frac{{16}}{{81}}$$
Solution :
The min. face value is not less than 2 and max. face value is not greater than 5 if we get any of the numbers 2, 3, 4, 5, while total possible out comes are 1, 2, 3, 4, 5 and 6.
∴ In one thrown of die, prob. of getting any no. Out of 2, 3, 4 and 5 is $$ = \frac{4}{6} = \frac{2}{3}$$
If the die is rolled four times, then all these events being independent, the required prob. $${\left( {\frac{2}{3}} \right)^4} = \frac{{16}}{{81}}$$