Question

1. Suppose that John and Tom are sitting in a classroom containing 9 students in total. A teacher randomly divides these 9 students into two groups: Group I with 4 students, Group II with 5 students (a) What is the probability that John is in Group 1? (b) If John is in Group I, what is the probability that Tom is also in Group 1? (c) What is the probability that John and Tom are in the same group? 2. In a certain group of people, it was found that 42% of them have alcoholic fathers, 8% of them have alcoholic mothers, and 48% of them have at least one alcoholic parent. If we randomly choose one individual from this group, what is the probability that: (a) the selected individual has two alcoholic parents? (b) the selected individual has an alcoholic mother but he/she does not have an alcoholic father? (c) the selected individual has an alcoholic mother, if he/she has an alcoholic father? (d) the selected individual has an alcoholic mother, if he/she does not have an alcoholic father?

Answer 1

The total number of students in the classroom are 9 Number of ways of selecting 4 students out of 9 for group I and 5 students for group II is (4) (5) 41(9-5)! = 126 5!(5-5)! The number of ways that John to be in the first group (3) (5) 3(813 , 5 (5-5), 56 The required probability that john is in the first group is P (John is in first group0.4444 56 126 Number of ways that John is in the first group is 56 Number of ways that that John and Tom in first group is 7(57! 2人5丿 2151 - 21 So, the required probability is 21 Probability - 56 - 0.375

Number of groups with John and Tom in first group is 7(5 7! 252!5! -21 Number of groups with John and Tom in second group is 7(3 7! -35 So, the number of groups such that John and Tom are in the same group is 21 +35-56 The required probability is, 56 126 -0.4444 P(John and Tom in same group) 2. Let A denote alcoholic fathers Let B denote alcoholic mothers The given values are as follows P(A) 0.42, P(B) 0.08, P(AUB) 0.48 It is required to find the probability that the selected individual has two alcoholic parents This implies 0.42+0.08-0.48 0.02 Hence, the probability that the selected individual has two alcoholic parents is 2%.

The probability that the selected individual has an alcoholic mother, but he/she does not have an alcoholic father is, 0.48-0.42 0.06 Hence, the probability that the selected individual has an alcoholic mother, but he/she does not have an alcoholic father is 6%. Compute the selected individual has an alcoholic mother, if he/she has an alcoholic father? P(B/A) 0.02 0.42 = 0.0476 This implies, 4.76% of the selected individual has an alcoholic mother, if he/she has an alcoholic father Compute the selected individual has an alcoholic mother, if he/she does not have an alcoholic father? P(A)-P(AnB) 0.08-0.02 1-0.42 0.103 Hence, 10.3% of the selected individual has an alcoholic mother. if he/she does not have an alcoholic father