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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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An unbiased die with faces marked 1,2,3,4,5 and 6 is rolled four times out of four 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 :

$\begin {array} {1 1} (A)\;\large\frac{16}{81} & \quad (B)\;\large\frac{1}{81} \\ (C)\;\large\frac{80}{81} & \quad (D)\;\large\frac{65}{81} \end {array}$


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Favourable points are 2,3,4,5
$ \therefore P = \large\frac{4}{6} = \large\frac{2}{3}$
Since the die is rolles 4 times
$ \therefore $ Required probability = $ \bigg( \large\frac{2}{3} \bigg)^4$
$ = \large\frac{16}{81}$
answered Jan 8, 2014 by thanvigandhi_1

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