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]$\frac{3}{20}$[/tex]

C. [tex]$\frac{2}{3}$[/tex]

D. [tex]$\frac{2}{20}$[/tex]



Answer :

To find the probability that both you and your friend are chosen as contestants from the 6 people in the front row, let's follow these steps:

1. Determine the total number of ways to choose 3 people from the 6 people:

The total number of ways to choose 3 people out of 6 is calculated using the combination formula:
[tex]\[ {}_6C_3 = \frac{6!}{3!(6-3)!} = 20 \][/tex]
Therefore, there are 20 ways to choose 3 people out of 6.

2. Determine the number of favorable ways where you and your friend are chosen:

If you and your friend are already chosen, we need to select 1 more person from the remaining 4 people. The number of ways to choose 1 person from 4 is calculated using:
[tex]\[ {}_4C_1 = \frac{4!}{1!(4-1)!} = 4 \][/tex]
So, there are 4 ways to choose the remaining 1 person from the 4 people after you and your friend have already been selected.

3. Calculate the probability:

The probability that both you and your friend are chosen is the ratio of the number of favorable outcomes to the total number of outcomes. Hence, the probability \( P \) is:
[tex]\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = 0.2 \][/tex]

So, the answer is [tex]\( A. \frac{4}{20} \)[/tex].