Question

The governing group of student-athletes on campus consists of six elected representatives from the various teams. Two of this year's group are female student-athletes. The athletic director plans to randomly select a subcommittee of 3 members to study and make recommendations about adding field hockey as a women's sport. The composition of the subcommittee is of some interest. The six students are Adam, Bob, Christopher, Doug, Elsa, and Frannie.

a. List all the simple events in this experiment. (Example: ABC, ABD, ABE etc)

b. Compute the probabilities of the following events of interest:

i: {both females are selected}

ii: {At least one female is selected}

iii: {No female is selected}

Answer 1

The probability of selecting both females for the subcommittee is 4/20, at least one female is 16/20, and no female is 4/20. These are based on the list of all 20 possible 3-person subcommittees from the six students.

Step-by-step explanation:

The question requires us to list all the possible combinations of a 3-person subcommittee and calculate the probability of specific compositions regarding the number of females on that committee. Given the six students: Adam, Bob, Christopher, Doug, Elsa, and Frannie, where Elsa and Frannie are the females.

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF.

• Both females are selected (EF combination with one of the four males): AEF, BEF, CEF, DEF = 4 events
• At least one female is selected: This includes all combinations except those with only males (ABC, ABD, ACD, BCD). Hence, it is 20 - 4 = 16 events.
• No female is selected: Only the combinations with all males: ABC, ABD, ACD, BCD = 4 events

To calculate the respective probabilities:

• Total possible subcommittees: 20 (from simple events)
• P(both females are selected) = 4/20
• P(at least one female is selected) = 16/20
• P(no female is selected) = 4/20

Note that these probabilities are calculated assuming that subcommittees of 3 are chosen at random from the 6 members, which constitutes a hypergeometric distribution.