Question

A lot consists of $144$ ball pens of which $20$ are defective and others are good. Nuri will buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it (ii) She will not buy it.

Answer 1

Given that total number of pens, say, $n\left(S\right)=144$
Given that total number of defective pen, $n\left(E\right)=20$
That means total number of non-defective pens, $n\left(\overline{E}\right)=144-20=124$

Solution(i):

For Nuri to buy a pen, it should be defective. Let the event of buying be $B$

No. of favourable outcomes $=$ No. of defective pens $=n\left(E\right)=20$

Total No. of outcomes $=$ total number of pens $=n\left(S\right)=144$

We know that, $P\left(B\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{n\left(E\right)}{n\left(S\right)}$$=\frac{20}{144}$ $=\frac{5}{36}$

Solution(ii):

For Nuri to not buy a pen, it should be non-defective. Let the event of not buying be $\overline{B}$

Since buying and not buying are the only events. The sum of probabilities of an event and its complement is $1$. i.e..,

$P\left(B\right)+P\left(\overline{B}\right)=1$

$P\left(\overline{B}\right)=1-P\left(B\right)$

$P\left(\overline{B}\right)=1-\frac{5}{36}$

$P\left(\overline{B}\right)=\frac{(36-5)}{36}$

$P\left(\overline{B}\right)=\frac{31}{36}$