We can use conditional probability to solve this problem:

1. Let P(A) be the probability of selecting a Type A widget, which is given as 0.6.
2. Let P(B) be the probability of selecting a Type B widget, which is 1 - P(A) = 1 - 0.6 = 0.4.
3. Let P(D) be the probability of selecting a defective widget, which is given as 0.05.
4. Let P(D|A) be the probability of a Type A widget being defective, which is given as 0.03.
5. We want to find P(D|B), the probability of a Type B widget being defective.
6. Using the formula for conditional probability, we have P(D) = P(D|A) * P(A) + P(D|B) * P(B).
7. Plugging in the given values, we get 0.05 = (0.03 * 0.6) + (P(D|B) * 0.4).

Solving for P(D|B), we find that P(D|B) = (0.05 - 0.018) / 0.4 = 0.032/0.4 = 0.08.

The probability of a Type B widget being defective is 0.08.
A factory produces two types of widgets: Type A and Type B. The probability of a randomly selected widget being Type A is 0.6, and the probability of a widget being defective is 0.05. If the probability of a Type A widget being defective is 0.03, what is the probability of a Type B widget being defective?