From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random.

Question

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If va...

Answer: 71

Accepted Answer

Given a lot of 12 items, 3 are defective.

Good items, $12-3=9$

Let $X$ denote the number of defective items.

So, value of $X=0,1,2,3$

A sample of $S$ items is drawn.

$P(X=0)=GGGGG$

(here $G$ is good item and $d$ is defective)

$\begin{gathered} \frac{9}{12}\cdot \frac{8}{11}\cdot \frac{7}{10}\cdot \frac{6}{9}\cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \ P(X=1)=5[\frac{9\cdot 8\cdot 7\cdot 6\cdot 3}{12\cdot 11\cdot 10\cdot 9\cdot 8}]=\frac{21}{44} \ P(X=2)=5[\frac{9\cdot 8\cdot 7\cdot 3\cdot 2}{12\cdot 11\cdot 10\cdot 9\cdot 8}]=\frac{14}{44} \ P(X=3)=5[\frac{3\cdot 2\cdot 1\cdot 9\cdot 8}{12\cdot 11\cdot 10\cdot 9\cdot 8}]=\frac{2}{44} \ P(X=4)=0 \ P(X=5)=0 \end{gathered}$

$X$
 0
 1
 2
 3
 4
 5

$P(X)$
 $\frac{7}{44}$
 $\frac{21}{44}$
 $\frac{14}{44}$
 $\frac{2}{44}$
 0
 0

$XP(X)$
 0
 $\frac{21}{44}$
 $\frac{28}{44}$
 $\frac{6}{44}$
 0
 0

${X}^{2}P(X)$
 0
 $\frac{21}{44}$
 $\frac{56}{44}$
 $\frac{18}{44}$
 0
 0

$\begin{gathered} {\sigma }_{2}^{x}=\sum {X}^{2}P(x)-{(\sum xP(x))}^{2} \ =\frac{95}{44}-{(\frac{55}{44})}^{2} \ =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \ =n-m=71 \end{gathered}$

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