**Question **: A die is thrown a fixed number of times. If probability of getting number 3 times is same as the probability of getting even number 4 times, then probability of getting even number exactly once is

(A) 1/6

(B) 1/9

(C) 5/36

(D) 7/128

**Answer **: option (D)

**Solution **:

According to the given condition,

nC_{3}\left( \dfrac{1}{2}\right) ^{n}=nC_{4}\left( \dfrac{1}{2}\right) ^{n}Where n is the number of times die is thrown

therefore, nC_{3}=nC_{1}

we get n=7.

Thus the required probability is,

7C_{1}\left( \dfrac{1}{2}\right) ^{7}=\dfrac{7}{2^{7}}=\dfrac{7}{128}