Question

A lot of 24 bulbs contains 25% defective bulbs. A bulb is drawn at random from the lot. It is found to be not defective and it is not put back. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

• A. $\frac{18}{23}$
• B. $\frac{20}{23}$
• C. $\frac{15}{24}$
• D. $\frac{17}{23}$

Answer 1

The correct option is D $\frac{17}{23}$

Correct Option is (C) 17/23

25% of 24 = $\frac{25}{100}$× 24 = 6.

So, there are 6 defective bulbs and 18 bulbs are not defective.

After the first draw, the lot is left with 6 defective bulbs and 17 non-defective bulbs.

So, when the second bulb is drwn, the total number of possible outcomes = 23 (= 6+ 17).

Number of favourable outcomes for the event E = number of non-defective bulbs = 17.

So, the required probability = P(E) = $\frac{17}{23}$.