Points: 0 of 10 A quality control inspector is checking a sample of lightbulbs for defects. The table summarizes the results. If one of these lightbulbs is selected at random, find the probability that the lightbulb is defective, given that it is 100 watts. \begin{tabular}{|r|r|r|r|} \hline Wattage & Good & Defective & Total \\ \hline 20 & 63 & 15 & 78 \\ 50 & 61 & 13 & 74 \\ 100 & 72 & 14 & 86 \\ Total & 196 & 42 & 238 \\ \hline \end{tabular} The probability is (Simplify your answer. Type an integer or a fraction.)

Step 1 ：The problem is asking for the conditional probability that a lightbulb is defective given that it is 100 watts. This can be calculated using the formula for conditional probability, which is \(P(A|B) = \frac{P(A \cap B)}{P(B)}\). In this case, event A is the lightbulb being defective and event B is the lightbulb being 100 watts.

Step 2 ：\(P(A \cap B)\) is the probability of both events happening, which is the number of 100 watt defective lightbulbs divided by the total number of lightbulbs. So, \(P(A \cap B) = \frac{14}{238} = 0.058823529411764705\).

Step 3 ：\(P(B)\) is the probability of the lightbulb being 100 watts, which is the total number of 100 watt lightbulbs divided by the total number of lightbulbs. So, \(P(B) = \frac{86}{238} = 0.36134453781512604\).

Step 4 ：Substitute \(P(A \cap B)\) and \(P(B)\) into the formula, we get \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.058823529411764705}{0.36134453781512604} = 0.16279069767441862\).

Step 5 ：Final Answer: The probability that a lightbulb is defective, given that it is 100 watts is \(\boxed{\frac{14}{86}}\) or approximately \(\boxed{0.163}\).