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

The correct answer is option D.  6/56

Step-by-step 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

~apex

Step-by-step explanation:

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.