A bag contains 5 defective pens and 7 working ones. A boy then a girl must each take a pen. Find the probability that at least one takes a defective pen?

Using the multiplication rule, we can find the probability of each event as follows: P(A) = 5/12, since there are 5 defective pens out of 12 total pens in the bag. P(B|A') = 5/11, since there are 5 defective pens left out of 11 total pens after the boy has taken a pen, and we are assuming that the boy did not take a defective pen (denoted by A'). (Note that we are using conditional probability here, since the probability of the girl taking a defective pen depends on whether or not the boy took a defective pen.) Using the addition rule, we can find the probability of the union of the two events as follows: P(A or B) = P(A) + P(B|A') - P(A and B) (Note that we subtract P(A and B) since we have counted the probability of both events occurring together twice.) To find P(A and B), we can use the multiplication rule as follows: P(A and B) = P(A) * P(B|A') (Note that we are assuming that the events are independent, which may not be strictly true in this case.) Substituting the values we have found, we get: P(A or B) = 5/12 + 5/11 - (5/12 * 5/11) = 121/264 ≈ 0.682 Therefore, the probability that at least one person takes a defective pen is approximately 0.682.