Question

A box contain 12 items out of which 4 are defective. A persons selects 6 items from the box. The expected number of defective items out of his selected items is: 

(a) 2 

(b) 3 

(c) \( 3 / 2 \) 

(d) none of the above

Answer 1

Correct option is (d) none of the above

\(p = \frac4{12}= \frac13\)

\(q = 1 - p = 1 - \frac13 = \frac23\)

\(n= 6\)

\(P(X =r) =\, ^nC_r \,p^r\,q^{n -r}\)

\(= \,^6C_r\,p^r\,q^{6-r}\)     \((\because n = 6)\) 

X	0	1	2	3	4
P(X)	\(\left(\frac 23\right)^6\)	\(2.\left(\frac23\right)^5\)	\(\frac53\left(\frac23\right)^4\)	\(\frac{20}{27}\left(\frac23\right)^3\)	\(\frac5{27}\left(\frac23\right)^2\)

\(E(X) = \displaystyle\sum^4_{i = 0} x_i(P(x_i))\)

\(= 0.\left(\frac23\right)^6 + 1. 2\left(\frac23\right)^5 + 2. \left(\frac53\right)\left(\frac23\right)^4+ 3.\left(\frac{20}{27}\right)\left(\frac23\right)^3 + 4 . \frac5{27}.\left(\frac23\right)^2\)

\(= \frac{64}{243}+ \frac{160}{243}+ \frac{160}{243}+\frac{80}{243}\)

\(= \frac{464}{243}\)

\(= 1.9\)