Question

A shipment of $8$ similar microcomputers to a retail outlet contains $3$ that are defective. If aschool makes a random purchase of $2$ of these computers, find the probability distribution for the number of defectives? Also, find its mean and standard deviation.

Answer 1

There are $8-3=5$ working microcomputers. There are ${}^8{C}_2=\frac{8!}{6!2!}=\frac{8\times 7\times 6!}{6!2!}=28$ ways to select $2$ computers.

There are ${}^3{C_0}^5{C}_2=1\times \frac{5\times 4\times 3!}{3!\times 2}=0$ ways to select $0$ defective computers and $2$ working computers.

The probability to select $0$ defective computers is

$P=\frac{10}{28}\approx 0.3571$

There are ${}^3{C_1}^5{C}_1=3\times \frac{5\times 4!}{4!\times 1}=15$ways to select $1$ defective computers and $1$ working computers.

The probability to select $1$ defective computers is

$P=\frac{3}{28}\approx 0.1071$

The numbers of defective microcomputers	$0$	$1$	$2$
Probability	$\frac{10}{28}$	$\frac{15}{28}$	$\frac{3}{28}$
${\left(Thenumbersofdefectivemicrocomputers\right)}^2$	$0$	$1$	$4$

Mean$=0\times \frac{10}{28}+1\times \frac{15}{28}+2\times \frac{3}{28}=\frac{0+15+6}{28}=\frac{21}{28}=0.75$

Mean of the squares$=0\times \frac{10}{28}+1\times \frac{15}{28}+4\times \frac{3}{28}=\frac{0+15+6}{28}=\frac{27}{28}=0.9642$

Standard deviation$=\sqrt{meanofthesquares-{mean}^2}=\sqrt{\frac{3}{14}}\approx 0.4629$