Probability Question 103
Question: If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1, is
Options:
A) $ \frac{2}{3} $
B) $ \frac{4}{5} $
C) $ \frac{7}{8} $
D) $ \frac{15}{16} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have mean $ (X)=np=2 $ and variance $ (X)=npq=1 $
Therefore $ q=\frac{1}{2} $ or $ p=\frac{1}{2} $ and $ n=4 $ Thus $ p(X\ge 1)=1-p(X=0)=1-{}^{4}C_0{{( \frac{1}{2} )}^{4}}=\frac{15}{16} $ .
BETA
