Question

A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in the previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?

Answer 1

Hint: We will first find the probability of getting a defective bulb when selected as random. So, we will get from the formula $P\left( \text{defective bulb} \right)=\dfrac{\text{No}\text{. of defective bulbs}}{\text{total no}\text{. of bulbs}}$ . After this, we will subtract 1 bulb from the total 20 bulbs as it is told that it is not replaced. So, we will have a total 19bulbs. Then, we will subtract 4 bulbs from it to get total non-defective bulbs because in the previous case there are chances that bulb was non-defective. Soo, then using the formula $P\left( \text{non-defective bulb} \right)=\dfrac{\text{No}\text{. of non-defective bulbs}}{\text{total no}\text{. of bulbs}}$ we will find the answer.

Complete step by step answer:
Here, we have a total number of bulbs 20. Number of defective bulbs is 4.
So, we have to find the probability of getting a defective bulb when a bulb is selected at random. So, we can write it as
$P\left( \text{defective bulb} \right)=\dfrac{\text{No}\text{. of defective bulbs}}{\text{total no}\text{. of bulbs}}$
So, on substituting the values, we will get as
$P\left( \text{defective bulb} \right)=\dfrac{4}{20}=\dfrac{1}{5}$ ………………………………(1)
Now, it is given that in the above case if the bulb selected is not defective and is not replaced. So, we have remaining bulbs as $20-1=19$ . Still we have 4 defective bulbs in a lot of 19 bulbs.
So, to find the number of non-defective bulbs we will subtract defective bulbs from total 19 bulbs. Thus, we will get a total number of non-defective bulbs as $19-4=15$ .
So, the probability of getting non-defective bulb is given as
 $P\left( \text{non-defective bulb} \right)=\dfrac{\text{No}\text{. of non-defective bulbs}}{\text{total no}\text{. of bulbs}}$
On substituting the values, we get as
$P\left( \text{non-defective bulb} \right)=\dfrac{15}{19}$ ……………………….(2)
Thus, the probability of defective bulb is $\dfrac{1}{5}$ and that of non-defective bulb is $\dfrac{15}{19}$ .

Note: Generally, students make mistakes by considering that bulb is replaced every time. So, there will be change in answer. Here, if we consider that bulb is replaced then, we will have total number of non-defective bulb as $20-4=16$ and the probability of non-defective bulbs will be $P\left( \text{non-defective bulb} \right)=\dfrac{16}{20}=\dfrac{8}{10}=0.8$ which is wrong. So, first carefully read the question and then solve the problem to avoid mistakes.