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We are given that 60% of the students wear neither a ring nor a necklace, 20% wear a ring, and 30% wear a necklace. We will use these probabilities to find the probabilities of the two questions: (a) A randomly chosen student is wearing a ring or a necklace. (b) A randomly chosen student is wearing both a ring and a necklace.

A helpful way to visualize the problem is to use a Venn diagram. Draw two circles, one representing the students wearing rings (R) and the other representing the students wearing necklaces (N). The overlap between the two circles represents the students wearing both a ring and a necklace.

To find the probability that a student is wearing a ring or a necklace, we will use the formula: P(A) = P(R) + P(N) - P(R ∩ N) where P(A) is the probability of a student wearing a ring or a necklace, P(R) is the probability of a student wearing a ring, P(N) is the probability of a student wearing a necklace, and P(R ∩ N) is the probability of a student wearing both a ring and a necklace. We are given that P(R) = 0.20 and P(N) = 0.30. To find P(R ∩ N), we can use the fact that 60% of students wear neither a ring nor a necklace, so the sum of the probabilities of the three distinct groups (ring only, necklace only, and both) should be 1 - 0.60 = 0.40. Thus, we have: P(R) + P(N) - P(R ∩ N) = 0.40 0.20 + 0.30 - P(R ∩ N) = 0.40 0.50 - P(R ∩ N) = 0.40 Solving for P(R ∩ N), we find that P(R ∩ N) = 0.10. Now, we can plug the values back into the formula for P(A): P(A) = 0.20 + 0.30 - 0.10 P(A) = 0.40 So, the probability that a randomly chosen student is wearing a ring or a necklace is 0.40 or 40%.

We already found that the probability of a student wearing both a ring and a necklace is P(R ∩ N) = 0.10. So, the probability that a randomly chosen student is wearing both a ring and a necklace is 0.10 or 10%.