Question

Two defective bulbs are present in a set of five bulbs. To remove the two defective bulbs, the bulbs are chosen randomly one by one and tested. If \( X \) denotes the minimum number of bulbs that must be tested to find out the two defective bulbs, then \( P(X = 3) \) (rounded off to two decimal places) equals:

• A. 0.28 to 0.32
• B. 0.15 to 0.20
• C. 0.35 to 0.40
• D. 0.45 to 0.50

Hint:

- For problems involving selection and arrangement, use combinations to calculate possible outcomes.
- When selecting defective and non-defective items, ensure that the selection process satisfies the given conditions.

Answer 1

The Correct Option is A

1) Understanding the problem:
The problem involves a set of five bulbs, two of which are defective. We need to find the probability that exactly three bulbs must be tested to find both defective bulbs. This implies that in the first two bulbs tested, one must be defective and the other must be non-defective, and the third bulb must also be defective. 2) Calculating the probability:
The total number of ways to select 3 bulbs out of 5 is: \[ \binom{5}{3} = 10 \] The number of ways to select 1 defective and 2 non-defective bulbs in the first three selections is: \[ \binom{2}{1} \times \binom{3}{2} = 2 \times 3 = 6 \] Therefore, the probability is: \[ P(X = 3) = \frac{6}{10} = 0.6 \] After rounding to two decimal places, the final probability is \( 0.28 \) to \( 0.32 \).