Question

A manufacturer assures his customers that the probability of having defective item is 0.005.A sample of 1000 items was inspected.find the probability of having he following possible outcomes
a)Only one is defective
b)At most 2 defective
c)More than 3 defective

Answer 1

Let $k$ be the number of defective items.

$p=0.005$ - the probability of having defective item

$q=1-p=0.995$ - the probability of having not defective item

We will use Binomial Distribution Formula:

In our case, probability that exactly $m$ items are defective is

$P(k=m)=C(1000,m)p^mq^{1000-m}$

a) $P(k=1)$ - probability that only one item is defective

$P(k=1)=C(1000, 1)p^1q^{999}=1000\times 0.005\times 0.995^{999}\approx 0.0334$

Answer: $0.0334$

b) $P(k\leq 2)$ - probability that at most two items are defective (i.e. $k=0,1,2)$

$P(k=0)=C(1000,0)p^0q^{1000}=1\times 1\times 0.995^{1000}\approx0.0066$

$P(k=1)\approx0.0334$

$P(k=2)=C(1000,2)p^2q^{998}=\frac{1000\times 999}{2} \times 0.005^2\times 0.995^{998}\approx0.0839$

$P(k\leq 2)\approx 0.0066+0.0334+0.0839\approx 0.1240$

Answer: $0.1240$

c) $P(k>3)$ - probability that more than three items are defective

$P(k>3)=1-P(k\leq3)=1-(P(k\leq 2)+P(k=3))$

$P(k=3)=C(1000,3)p^3q^{997}=\frac{1000\times 999\times 998}{3\times 2\times 1}\times 0.005^3\times 0.995^{997}\approx0.1403$

$P(k>3)\approx 1-(0.1240+0.1403)\approx 0.7357$

Answer: $0.7357$