Question

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

Answer 1

The probability of selecting a junior or a female is 6/7,

the probability of selecting a senior or a female is 4/7, and

the probability of selecting a junior or a senior is 1 since all students fall into one of those two categories.

Step-by-step explanation:

To find the probability of selecting a junior or a female, we must consider all the juniors and all the females, taking care not to double-count the female juniors. There are 18 juniors and 6 senior females. Since 12 juniors are male, there are 18 - 12 = 6 female juniors. Combining the 6 female seniors and 6 female juniors, there are 12 females. So, the total number of juniors or females is 18 (juniors) + 12 (females) - 6 (female juniors already counted with juniors) = 24. Since there are 28 students in total (18 juniors + 10 seniors), the probability is 24/28 which simplifies to 6/7.

The probability of selecting a senior or a female involves counting all seniors and females. As calculated before, there are 12 females. Since no additional information has been provided, the 10 seniors include the 6 female seniors we already counted. Hence, the probability is (10 + 12 - 6) / 28 = 16/28 which simplifies to 4/7.

The probability of selecting a junior or a senior is simply the probability of selecting any student since the only students available are juniors and seniors. Therefore, the probability is 1 or 100%.

Answer 2

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.