In Exercises 16–18, use the following information. At a college, 43% of the students are women, 15% of the students are art majors, 4% have not chosen a major, and 8% are women and art majors. A faculty member conducts an experiment and selects students at random to participate.

16. What is the probability that a randomly selected student will be either a woman or an art major?
Would this one be 43/70 + 15/70 - 8/70 = 5/7
Or would the denominator be 100 and equal 1/2

43/100 women

15/100 art
take away the 8/100 that are both
yes I get 1/2

Bless you

You are welcome. I think they left out the 38 percent who majored in physics :)

(To get 70% you counted the 8 twice, I think really the total is 62 but they gave percent, not number of students so it is out of 100 )

To find the probability that a randomly selected student will be either a woman or an art major, you can use the principle of inclusion-exclusion.

Here's how to calculate it:

Step 1: Find the probability of selecting a woman.
Given that 43% of the students are women, the probability of selecting a woman is 43/100.

Step 2: Find the probability of selecting an art major.
Given that 15% of the students are art majors, the probability of selecting an art major is 15/100.

Step 3: Find the probability of selecting a student who is both a woman and an art major.
Given that 8% of the students are women and art majors, the probability of selecting a student who is both is 8/100.

Step 4: Apply the principle of inclusion-exclusion.
Use the formula: P(A or B) = P(A) + P(B) - P(A and B)

In this case, A represents the event of selecting a woman and B represents the event of selecting an art major. So, the probability you are looking for is:
P(woman or art major) = P(woman) + P(art major) - P(woman and art major)
= (43/100) + (15/100) - (8/100)
= 50/100
= 1/2

Therefore, the probability that a randomly selected student will be either a woman or an art major is 1/2, or 50%.