Question

There are 10 pens such that 3 pens are defective. Let X represent the number of defective pen selected. If two pens are selected at random then variance of X is

(1) \(\frac{38}{75}\)

(2) \(\frac{28}{75}\)

(3) \(\frac{14}{75}\)

(4) \(\frac{3}{75}\) 

Answer 1

Correct option is: (2) \(\frac{28}{75}\)  

\(\mu=\sum X_{i} P\left(X_{i}\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5}\)

Variance (X) = \(\sum P_{i}\left(X_{i}-\mu\right)^{2}=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{1}{15}\left(\frac{49}{25}\right)=\frac{28}{75}\)