Probability Question 396
Question: If 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{11}{16} $
Show Answer
Answer:
Correct Answer: D
Solution:
We hove; np =2=mean Npq=1=variance
$ \Rightarrow p=\frac{1}{2};q=\frac{1}{2}\And n=4 $ Required probability $ =P(x>1) $
$ =1-P(x\le 1) $
$ =1-[P(x=0)+P(x=1)] $
$ =1-{{[}^{4}}C_0q^{4}{{+}^{4}}C_1q^{3}p^{1}] $
$ =1-\frac{5}{16}=\frac{11}{16} $
BETA
