ash4gmat wrote: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.
Let A be the event that Mark's little brother selects an outfit with black slacks and B be the event that he selects an outfit with a blue shirt.
What is P(A or B), the probability that Mark's little brother selects an outfit with black slacks or an outfit with a blue shirt.

OA is 2/3

First, pick a shirt: 3 choices

Next, pick slacks: 2 choices

Finally, pick a tie: 1 choice

Total combinations: 3 x 2 x 1 = 6 outfits

These are split equally between the two pairs of slacks, so

3 outfits with black slacks out of 6 = 1/2

Similarly, the 6 outfits are split equally between the 3 shirt choices, so

2 outfits with blue shirt out of 6 = 1/3

Recognize that the outfits above BOTH include a combination of blue shirt and black slacks, so to simply add 1/2 and 1/3 would be double counting this combination. So, add 1/2 and 1/3 and subtract the probability of the blue shirt and black slack outfit, which is 1 of the 6 outfits:

1/2 + 1/3 -1/6 = 3/6 + 2/6 - 1/6 = 4/6 = 2/3

Expert solutions!!!

Let's think about what we CAN'T have: grey slacks AND a white or black shirt.

The probability of the grey slacks is (1/2), and the probability of either the white or black shirt is (2/3).

The probability of both of these (the horror!) is (1/2) * (2/3), or 1/3. So there's a (1/3) chance of getting what we DON'T want, meaning there's a 1 - 1/3, or 2/3 chance of getting what we DO want.

That said, we're trusting Mark's little brother to pick exactly one shirt and exactly one set of slacks, which seems like a lot of faith in little brothers.