Problem 10 - A First Course in Probability

Question

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace? (b) a ring and a necklace?

Step-by-Step Solution

Answer

The probability that a randomly chosen student is wearing (a) a ring or a necklace is \(0.40\) or \(40\%\) and (b) both a ring and a necklace is \(0.10\) or \(10\%\).

Step 1: Understand the question

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.

Step 2: Visualize the problem

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.

Step 3: Probability of a student wearing a ring or a necklace (A)

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%.

Step 4: Probability of a student wearing a ring and a necklace (B)

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%.