Question

A lot of $20$ bulbs contain $4$ defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Suppose the bulb drawn in $\mathrm{i}$ is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer 1

Step 1: Calculate the probability that the first bulb drawn is defective

It is given that the total number of bulbs in the lot $=20$

And, the number of defective bulbs $=4$

Now, as we know,

$\mathbf{Probability}\mathbf{}\mathbf{of}\mathbf{}\mathbf{an}\mathbf{}\mathbf{event}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}$

$\Rightarrow$ The probability that the first bulb drawn is defective $=\frac{\mathrm{Number}\mathrm{of}\mathrm{defective}\mathrm{bulbs}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{bulbs}}$

$\Rightarrow$ The probability that the first bulb drawn is defective $=\frac{4}{20}$

$\Rightarrow$ The probability that the first bulb drawn is defective $=\frac{1}{5}$

$\Rightarrow$ The probability that the first bulb drawn is defective $=0.2$

Hence, the probability that the first bulb drawn is defective is $0.2$.

Step 2: Calculate the probability that the second bulb drawn is not defective

According to the question, the bulb drawn in the subpart $\mathrm{i}$ is not defective and is not replaced.

So, the total number of bulbs available for the second draw $=19$

And, the number of defective bulbs $=4$

So, the number of non-defective bulbs $=19-4=15$

Again, as we know,

$\mathbf{Probability}\mathbf{}\mathbf{of}\mathbf{}\mathbf{an}\mathbf{}\mathbf{event}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}$

$\Rightarrow$ The probability that the second bulb drawn is not defective $=\frac{\mathrm{Number}\mathrm{of}\mathrm{non}-\mathrm{defective}\mathrm{bulbs}}{\mathrm{Number}\mathrm{of}\mathrm{bulbs}\mathrm{available}}$

$\Rightarrow$ The probability that the second bulb drawn is not defective $=\frac{15}{19}$

$\Rightarrow$ The probability that the second bulb drawn is not defective $=0.789$

Hence, the probability that the second bulb drawn is not defective is $0.789$.

The probability that the first bulb drawn is defective is $0.2$.

The probability that the second bulb drawn is not defective is $0.789$.