In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are females and 12 of the juniors are males. If a student is selected at random, find the probability of selecting the following.

a. A junior or a female

b. A senior or a female

c. A junior or a senior

a) 0.857

b) 0.571

c) 1

Step-by-step explanation:

Based on the data given, we have

18 juniors 10 seniors 6 female seniors 10-6 = 4 male seniors 12 junior males 18-12 = 6 junior female 6+6 = 12 female 4+12 = 16 male A total of 28 students

The probability of each union of events is obtained by summing the probabilities of the separated events and substracting the intersection. I will abbreviate female by F, junior by J, male by M, senior by S. We have

P (J U F) = P (J) + P (F) - P (JF) = 18/28+12/28-6/28 = 24/28 = 0.857 P (S U F) = P (S) + P (F) - P (SF) = 10/28 + 12/28 - 6/28 = 16/28 = 0.571 P (J U S) = P (J) + P (S) - P (JS) = 18/28 + 10/28 - 0 = 1

Note that a student cant be Junior and Senior at the same time, so the probability of the combined event is 0. The probability of the union is 1 because every student is either Junior or Senior.