Question

Lot of 100 items contain 10 defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability thet the store will receive at most one defective item?

Answer 1

Total no. of items $=100$

No. of defective items $=10$

No. of safe items $=100-10=90$

Let $p$ and $q$ be the probability of defective and safe items respectively.

$p=\frac{10}{100}=\frac{1}{10}$

$q=\frac{90}{100}=\frac{9}{10}$

$n=5$

As we know that,

$P(x=r)={}^n{C}_r{\left(p\right)}^r{\left(q\right)}^{n-r}$

Therefore,

$P(x\le 1)=P(x=0)+P(x=1)$

$P(x\le 1)={}^5{C}_0{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^5+{}^5{C}_1{\left(\frac{1}{10}\right)}^1{\left(\frac{9}{10}\right)}^4$

$P(x\le 1)={\left(\frac{9}{10}\right)}^4(\frac{9}{10}+\frac{5}{10})$

$\Rightarrow P(x\le 1)={\left(\frac{9}{10}\right)}^4\left(\frac{7}{5}\right)$

Hence the probability that the store will receive at most one defective item is ${\left(\frac{9}{10}\right)}^4\left(\frac{7}{5}\right)$.

Hence the correct answer is ${\left(\frac{9}{10}\right)}^4\left(\frac{7}{5}\right)$.