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

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

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,

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