At a game show, there are 8 people (including you and your friend) in the front row. The host randomly chooses 3 people from the front row to be contestants. The order in which they are chosen does not matter. What is the probability in fraction form that you and your friend are both chosen?
There are 8 people in the front row, so there are a total of 8 choose 3 ways to choose 3 people from the front row.
The number of ways to choose you and your friend is 1 choose 1 (you) multiplied by 1 choose 1 (your friend), which is equal to 1.
Therefore, the probability that you and your friend are both chosen is 1/(8 choose 3).
Calculating this fraction:
8 choose 3 = 8! / (3!(8-3)!) = 8! / (3!5!) = (8*7*6*5*4*3*2*1) / (3*2*1)(5*4*3*2*1) = (8*7*6) / (3*2*1) = 56
So, the probability is 1/56.