Probability Question 147
Question: The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
[AIEEE 2004]
Options:
A) $ \frac{28}{256} $
B) $ \frac{219}{256} $
C) $ \frac{128}{256} $
D) $ \frac{37}{256} $
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Answer:
Correct Answer: A
Solution:
$ . \begin{vmatrix} np=4 \\ npq=2 \\ \end{vmatrix} \ \Rightarrow q=\frac{1}{2},p=\frac{1}{2},p=\frac{1}{2},n=8 $
$ P(X=2)={{}^{8}}C_2{{( \frac{1}{2} )}^{2}}{{( \frac{1}{2} )}^{6}}=28.\frac{1}{2^{8}}=\frac{28}{256} $ .
BETA
