- Home
- Standard 11
- Mathematics
14.Probability
medium
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
$ 1/81 $
C
$80/81$
D
$65/81$
(IIT-1993)
Solution
(a) $P($ minimum face value not less than $2$ and maximum face value is not greater than $5$)
$ = P(2\,{\rm{or}}\,3\,{\rm{or}}\,4\,{\rm{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}}.$
Standard 11
Mathematics
