Probability Question 304
Question: The mean and the variance of binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
Options:
A) $ \frac{28}{256} $
B) $ \frac{219}{256} $
C) $ \frac{128}{256} $
D) $ \frac{37}{256} $
Show Answer
Answer:
Correct Answer: A
Solution:
Mean $ =np=4 $ and variance $ =npq=2 $
$ \therefore p=q=\frac{1}{2}andn=8 $
$ \therefore $ P (2 success) $ ={{}^{8}}C_2{{( \frac{1}{2} )}^{6}}{{( \frac{1}{2} )}^{2}}=\frac{28}{2^{8}}=\frac{28}{256} $
BETA
