four fair dice `D_1, D_2,D_3 and D_4` each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that `D_4` shows a number appearing on one of `D_1,D_2,D_3` is

Options

A. `(91)/(216)`
B. `(108)/(216)`
C. `(125)/(216)`
D. `(127)/(216)`

Answer

A

Explanation

Sample space `=6xx6xx6xx6=6^(4)`
Favourable cases = Case I or Case II or Casr III
Case I First, we should select one number for `D_(4)` which appears on all i.e., `.^(6)C_(1)xx1`.
Case II For `D_(4)`, there are `.^(6)C_(1)` ways. Now, it appears on any one of `D_(1), D_(2), D_(3)` i.e., `.^(3)C_(1)xx1`.
For other two there are `5xx5` ways
Total number of ways `.^(6)C_(1)xx .^(3)C_(1)xx 1xx5xx5`
Casr III For `D_(4)`, there are `.^(6)C_(1)` ways. Now, it appears on any two of `D_(1),D_(2), D_(3) rArr .^(3)C_(2)xx1^(2)`
For other one there are 5 ways.
Total number of ways `= .^(6)C_(1)xx.^(3)C_(2)xx1^(2)xx5`
`therefore` Required probability
`=(.^(6)C_(1)+ .^(6)C_(1)xx .^(3)C_(1)xx 5^(2)+ .^(6)C_(1)xx .^(3)C_(2)xx5)/(6^(4))`
`=(6(1+75+15))/(6^(4))=(91)/(216)`