Question

In a statistics class, there are 22 juniors and 18 seniors. Of the juiors, 12 are females. Of the seniors, 11 are males. What is the probability that a randomly selected studemt from the class is a male or a junior?

Answer 1

Answer:

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.