At a game show, there are 6 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 [tex]${ }_6 C_3=20$[/tex] total ways to choose the 3 contestants.

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

A. [tex]\frac{4}{20}[/tex]

B. [tex]\frac{2}{3}[/tex]

C. [tex]\frac{2}{20}[/tex]

D. [tex]\frac{3}{20}[/tex]



Answer :

To determine the probability that both you and your friend are chosen as contestants out of the 6 people, while knowing that the host will choose 3 people in total, let's break down the problem step by step:

1. Total Number of Ways to Choose 3 Out of 6 People:
The total number of ways to choose 3 people from 6 is given by the combination formula [tex]\({}_6C_3\)[/tex]:
[tex]\[ {}_6C_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} = 20 \][/tex]

2. Number of Favorable Outcomes:
We need to find the number of ways to choose 3 people such that both you and your friend are included among the chosen ones.

- Since you and your friend must be chosen, we are left with choosing 1 more person out of the remaining 4 people.
- The number of ways to choose 1 person out of 4 is given by the combination formula [tex]\({}_4C_1\)[/tex]:
[tex]\[ {}_4C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1! \cdot 3!} = 4 \][/tex]

3. Probability Calculation:
The probability is the ratio of the number of favorable outcomes to the total number of outcomes.
[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{{}_4C_1}{{}_6C_3} = \frac{4}{20} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{\frac{4}{20}} \][/tex]