A die is thrown a fixed number of times. If probability of getting even number 3 times is same as the probability of getting even number 4 times, then probability of getting even number exactly once is `1//6` b. `1//9` c. `5//36` d. `7//128`

Options

A. `1//6`
B. `1//9
C. `5//36`
D. `7//128`

Answer

D

Explanation

According to the givn condition,
`""^(n)C_(3)((1)/(2))^(n)=""^(n)C_(4)((1)/(2))^(n),`
where n is the number of times die is thrown.
`therefore""^(n)C_(3)=""^(n)C_(4)impliesn=7`
Thus, the required probability is
`=""^(7)C_(1)((1)/(2))^(7)=(7)/(2^(7))=(7)/(128)`