Let's determine the probability that you and your friend are both chosen as contestants from the group of 8 people.
First, we need to recognize that you want to ensure that you and your friend are already chosen, so we are looking at a specific scenario.
There are a total of 8 people in the row, and we need to choose 3 out of these 8 to be contestants. We are given that there are a total of [tex]\({}_8 C_3 = 56\)[/tex] ways to choose any 3 people out of the 8.
Next, we need to consider the specific situation where you and your friend are two of the chosen contestants. We are left to choose just one more contestant from the remaining 6 people.
The number of ways to choose this remaining third contestant from the remaining 6 people is given by:
[tex]\[
{}_6 C_1
\][/tex]
Calculating [tex]\({}_6 C_1\)[/tex]:
[tex]\[
{}_6 C_1 = 6
\][/tex]
Now, we need the probability that you and your friend are included in the 3 chosen contestants. Since there are 6 ways to pick 1 additional contestant to join you and your friend in the game show, and there are 56 ways to choose any 3 contestants from the original 8, the probability is:
[tex]\[
\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{6}{56}
\][/tex]
So, the probability that both you and your friend are chosen is:
[tex]\[
\frac{6}{56}
\][/tex]
Thus, the correct answer is:
D. [tex]\(\frac{6}{56}\)[/tex]
