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}