Question 6 ( 2 marks) In a group of 50 people there are 30 who are on the on the athletics team and 27 who play basketball. Twelve of the people on the athletics team also play basketball and five people in the class are involved in neither. A person from the group is chosen at random. (a) What is the probability that the person is on the athletics team and plays basketball? b) What is the probability that the person is on the athletics team or plays basketball, or both?

To solve this problem, we will use the principles of set theory and probability. ### Known Conditions: - Total number of people in the group, \( N = 50 \) - Number of people on the athletics team, \( A = 30 \) - Number of people who play basketball, \( B = 27 \) - Number of people on both teams, \( A \cap B = 12 \) - Number of people involved in neither, \( N_{\text{neither}} = 5 \) ### Step-by-Step Solution: #### (a) Probability that the person is on the athletics team and plays basketball The probability \( P(A \cap B) \) can be calculated using the formula: \[ P(A \cap B) = \frac{N(A \cap B)}{N} \] Where: - \( N(A \cap B) = 12 \) (people on both teams) - \( N = 50 \) (total people) Now, substituting the values: \[ P(A \cap B) = \frac{12}{50} \] Calculating this gives: \[ P(A \cap B) = 0.24 \] #### (b) Probability that the person is on the athletics team or plays basketball, or both To find the probability \( P(A \cup B) \), we can use the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \( P(A) = \frac{N(A)}{N} = \frac{30}{50} = 0.6 \) - \( P(B) = \frac{N(B)}{N} = \frac{27}{50} = 0.54 \) - \( P(A \cap B) = 0.24 \) (calculated in part a) Now substituting these values into the formula: \[ P(A \cup B) = 0.6 + 0.54 - 0.24 \] Calculating this gives: \[ P(A \cup B) = 0.6 + 0.54 - 0.24 = 0.9 \] ### Final Answers: (a) The probability that the person is on the athletics team and plays basketball is \( 0.24 \). (b) The probability that the person is on the athletics team or plays basketball, or both is \( 0.9 \).