Question

Mark is going to an awards dinner and wants to dress appropriately. He is running behind schedule and asks his little brother to randomly select an outfit for him.

Mark has one blue dress shirt, one white dress shirt, one black dress shirt, one pair of black slacks, one pair of grey slacks, and one red tie. All six of his possible outfits are listed below.
Let
A
AA be the event that Mark's little brother selects an outfit with a white shirt and grey slacks and
B
BB be the event that he selects an outfit with a black shirt.
What is
P
(
A
or
B
)
P(A or B)P, left parenthesis, A, start text, space, o, r, space, end text, B, right parenthesis, the probability that Mark's little brother selects an outfit with a white shirt and grey slacks or an outfit with a black shirt?

Answer 1

There are six possible outfits, and we want to find the probability that Mark's little brother selects an outfit with a white shirt and grey slacks or an outfit with a black shirt. We can use the addition rule of probability:

P(A or B) = P(A) + P(B) - P(A and B)

where P(A) is the probability of selecting an outfit with a white shirt and grey slacks, P(B) is the probability of selecting an outfit with a black shirt, and P(A and B) is the probability of selecting an outfit with both a white shirt and grey slacks, and a black shirt.

From the six possible outfits, there is one outfit with a white shirt and grey slacks, and one outfit with a black shirt, so:

P(A) = 1/6
P(B) = 1/6

There is no outfit that satisfies both events A and B, so:

P(A and B) = 0

Therefore:

P(A or B) = P(A) + P(B) - P(A and B) = 1/6 + 1/6 - 0 = 1/3

So the probability that Mark's little brother selects an outfit with a white shirt and grey slacks or an outfit with a black shirt is 1/3.