Question

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.

There are ^8C^3 = 56 total ways to choose the 3 contestants.

What is the probability that you and your friend are both chosen?

A. 2/56
B. 3/56
C. 2/3
D. 6/56

Answer 1

`(6)/(56)` ~apex

Explanation:

Answer 2

The correct answer is option D. 6/56

Explanation:

It is given that, there are 8 people.

The host randomly chooses 3 people from the front row to be contestants.

To find the required probability

There are 56 ways to choose the 3 contestants.

2 people must be selected. Remaining 1 is selected from 6 people in 6 ways.

Therefore required probability = 6/56

The correct answer is option D. 6/56