Question 293 of 417EasyMCQ
The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P(X=1) is : Solutions · 1
Official solution
Mean = np = 4
Variance = npq = 2
\[ \Rightarrow 4 \times q\] = 2
\[ \Rightarrow\] q = \[{1 \over 2}\]
p = 1 - q = 1 - \[{1 \over 2}\] = \[{1 \over 2}\]
n = 8
Now P(X = 1) = \[{}^8{C_1}{\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{8 - 1}}\]
= \[8 \times {1 \over {{2^8}}}\]
= \[{1 \over {32}}\] Answer the question to unlock