A die is thrown `31` times. The probability of getting `2,4` or `5` at most `15` times is

Options

A. `(1)/(3)`
B. `(1)/(4)`
C. `(1)/(5)`
D. `(1)/(2)`

Answer

D

Explanation

`(d)` The probability of getting `2` or `4` or `5` in a single trial is
`p=(1)/(6)+(1)/(6)+(1)/(6)=(1)/(2)`
`:.q=(1)/(2)`
The probability of getting `2` or `4` or `5` at most `15` times while throwing a die `31` times
`="^(31)C_(0)*q^(31)+^(31)C_(1)*pq^(30)+^(31)C_(2)p^(2)q^(29)+....+^(31)C_(15)p^(15)q^(16)`
`=("^(31)C_(0)+^(31)C_(1)+....+^(31)C_(15))(1)/(2^(31)) ( :' p=q=(1)/(2))`
`=(1)/(2)("^(31)C_(0)+^(31)C_(1)+....+^(31)C_(31))(1)/(2^(31))` ( `:' ^(31)C_(16)=^(31)C_(15)` etc.)
`=(1)/(2)*2^(31)*(1)/(2^(31))=(1)/(2)`