If a light bulb is selected at random from all the defective bulbs produced at the factory, what is the probability tha…

To solve this problem, we can assume a convenient total number of bulbs produced that reflects the 7:3 ratio, such as 1,000 bulbs.

Step 1: Calculate the number of each type of bulb produced.

The ratio of LED to Halogen bulbs is 7:3, which means for every 10 bulbs, 7 are LED and 3 are Halogen.

Number of LED bulbs = (7 / (7+3)) · 1000 = (7/10) · 1000 = 700.

Number of Halogen bulbs = (3 / (7+3)) · 1000 = (3/10) · 1000 = 300.

Step 2: Calculate the number of defective bulbs for each type.

The defect rate for LED is 4%, and for Halogen is 8%.

Number of defective LED bulbs = 0.04 · 700 = 28.

Number of defective Halogen bulbs = 0.08 · 300 = 24.

Step 3: Calculate the total number of defective bulbs.

Total defective bulbs = 28 (defective LED) + 24 (defective Halogen) = 52.

Step 4: Calculate the conditional probability.

The question asks for the probability that a randomly selected defective bulb is an LED bulb. This is the number of defective LED bulbs divided by the total number of defective bulbs.

Probability = (Number of defective LED bulbs) / (Total number of defective bulbs) = 28 / 52.

Step 5: Simplify the fraction.

28/52 can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4.

28 ÷ 4 = 7.

52 ÷ 4 = 13.

The probability is 7/13.

Therefore, the correct answer is A.

B is incorrect. This is the probability that a randomly selected defective bulb is a Halogen bulb (24/52 = 6/13).

C is incorrect. This is the overall probability of selecting an LED bulb from the total production (700/1000 = 7/10), without considering that the selected bulb must be defective.

D is incorrect. This is the probability that a randomly selected LED bulb is defective (4% = 4/100 = 1/25). This reverses the condition of the problem, which selects from the population of defective bulbs.