Question

Suppose that a box contains 25 bulbs, of which 20 are good and the other 5 are defective. Consider randoml selecting three bulbs without replacement. Let \(E\) denote the event that the first bulb selected is good, \(F\) be the event that the second bulb is good, and \(G\) represent the event that the third bulb selected is good. a. What is \(P(E)\) ? b. What is \(P(F \mid E)\) ? c. What is \(P(G \mid E \cap F)\) ? d. What is the probability that all three selected bulbs are good?

Short answer

a. \(P(E) = 0.8\) \n b. \(P(F | E) = 0.792\) \n c. \(P(G | E ∩ F) = 0.783\) \n d. The probability that all three selected bulbs are good is 0.496.

Step-by-step solution

Compute P(E)

This is the probability of picking a good bulb on the first try. There are 20 good bulbs out of the total of 25, so \( P(E) = \frac{20}{25} = 0.8 \)

Compute P(F | E)

This is the probability of picking a good bulb on the second try given that a good bulb was selected first. After picking the first bulb which was good, there are 19 good bulbs left and 24 bulbs in total. Therefore, \( P(F | E) = \frac{19}{24} = 0.792 \)

Compute P(G | E ∩ F)

This is the probability of picking a good bulb on the third try given that the first two selections were good bulbs. After picking two bulbs which were good, there are 18 good bulbs left and 23 bulbs in total. Therefore, \( P(G | E ∩ F) = \frac{18}{23} = 0.783 \).

Compute the probability that all three selected bulbs are good

The probability that all three bulbs are good can be calculated by multiplying the probabilities of E, F and G. Therefore, \( P(E ∩ F ∩ G) = P(E) × P(F | E) × P(G | E ∩ F) = 0.8 × 0.792 × 0.783 = 0.496 \).