Question

There are $5$ defective items in a large bulk of $100$ items. What is the probability that a sample of $10$ items will include not more than one defective item?

• A.
  ${\left(\frac{19}{20}\right)}^{10}$
• B.
  ${\left(\frac{9}{20}\right)}^9\frac{19}{20}$
• C.
  ${\left(\frac{19}{20}\right)}^9\frac{29}{20}$
• D. ${\left(\frac{9}{20}\right)}^{10}\frac{19}{20}$

Answer 1

The correct option is C

${\left(\frac{19}{20}\right)}^9\frac{29}{20}$
Let there be $X$ number of defectve items in a sample of ten items drawn successively.
Now, as we can see that the drawing of the items is done with replacement. Thus, the trials are Bernoulli trials.
Now, pobability of getting a defective item, $p=\frac{5}{100}=\frac{1}{20}$
Thus, $q=1-\frac{1}{20}=\frac{19}{20}$
$\therefore$ We can say that x has binomial distribution, where $n=10\text{and}p=\frac{1}{20}$
Thus, $P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$ where $x=0,1,2,...n$
${=}^{10}{C}_x{\left(\frac{19}{20}\right)}^{10-x}{\left(\frac{1}{20}\right)}^x$
Probability of getting not more than one defective item $=P(X\le 1)$
$=P(X=0)+P(X=1){=}^{10}{C}_0{\left(\frac{19}{20}\right)}^{10}{\left(\frac{1}{20}\right)}^0{+}^{10}{C}_1{\left(\frac{19}{20}\right)}^9{\left(\frac{1}{20}\right)}^1={\left(\frac{19}{20}\right)}^9[\frac{19}{20}+\frac{10}{20}]={\left(\frac{19}{20}\right)}^9\times \left(\frac{29}{20}\right)$