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: Firstly find the possible number of outcomes P(Y) that is the total number of bulbs that can be drawn from the lot and wanted number of outcomes P(X) that is the defective number of bulbs that can be drawn from the lot. Calculate the probability by using the formula, probability of an event P(E) is given as $P\left( E \right)=\dfrac{P\left( X \right)}{P\left( Y \right)}$. Similarly calculate the probability for the second case.

Complete step by step solution:
We know the probability of any event P(E) is the ratio of number of wanted outcome P(X) to the number of possible outcomes P(Y) and is given by the formula $P\left( E \right)=\dfrac{P\left( X \right)}{P\left( Y \right)}$.
We know the total number of bulbs in the lot is 20, therefore we can say that the possible number of outcomes P(Y) is 20.
Similarly 4 bulbs out of 20 are defective ones hence the wanted number of outcomes P(X) is 4.
Applying the above mentioned formula to find out the probability of getting a defective bulb P(D),
$P\left( D \right)=\dfrac{P\left( X \right)}{P\left( Y \right)}$
By substituting the values of P(X) and P(Y) in the above expression we get,
$P\left( D \right)=\dfrac{4}{20}$
On simplify the above expression the probability of getting a defective bulb P(D) is,
$P\left( D \right)=\dfrac{1}{5}$
Hence we get the probability of getting a defective bulb $P\left( D \right)=\dfrac{1}{5}$.
If the bulb selected in the previous case is not defective and not replaced then the possible number of outcomes P(B) reduces to 19 from 20.
If the bulb selected in the previous case is not defective and not replaced then the wanted number of outcomes P(A) reduces to 15 from 16.
Applying the probability formula to find out the probability of getting a non-defective bulb P(N),
$P\left( N \right)=\dfrac{P\left( A \right)}{P\left( B \right)}$
By substituting the values of P(A) and P(B) in the above expression we get,
$P\left( N \right)=\dfrac{15}{19}$
Hence we get the probability of getting a non-defective bulb $P\left( N \right)=\dfrac{15}{19}$.

Note: Alternatively the probability of getting a non-defective bulb P(N) can also be calculated by subtracting the probability of getting a defective bulb P(D) from 1 after the possible number of outcomes become 19 and P(D) is also improvised according to the same data. A common error that you may encounter could be that the possible number of outcomes are not updated after the first random selection of bulbs.