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 [tex]${ }_8 C_3=56$[/tex] total ways to choose the 3 contestants.

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

A. [tex]$\frac{6}{56}$[/tex]
B. [tex][tex]$\frac{2}{3}$[/tex][/tex]
C. [tex]$\frac{3}{56}$[/tex]
D. [tex]$\frac{2}{56}$[/tex]



Answer :

To determine the probability that both you and your friend are chosen as contestants from the group of 8 people, we need to perform several steps. Here's the detailed, step-by-step solution:

### Step 1: Calculate the Total Number of Ways to Choose 3 People from 8
The total number of ways to choose 3 people out of 8 is given by the combination formula [tex]\( \binom{8}{3} \)[/tex]:
[tex]\[ \binom{8}{3} = 56 \][/tex]
So, there are 56 possible ways to choose 3 contestants from 8 people.

### Step 2: Calculate the Number of Favorable Ways
To find the number of favorable ways in which both you and your friend are chosen, we start by considering that you and your friend are already chosen. So, we need to choose the remaining contestant from the other 6 people.

The number of ways to choose 1 person from the remaining 6 people is given by the combination formula [tex]\( \binom{6}{1} \)[/tex]:
[tex]\[ \binom{6}{1} = 6 \][/tex]
Thus, there are 6 favorable ways to choose 1 additional contestant from the remaining 6 people, ensuring both you and your friend are chosen.

### Step 3: Calculate the Probability
The probability that both you and your friend are chosen is the ratio of the number of favorable ways to the total number of ways to choose 3 contestants from the group of 8 people:
[tex]\[ \text{Probability} = \frac{\text{Number of Favorable Ways}}{\text{Total Number of Ways}} = \frac{6}{56} \][/tex]

Simplifying the fraction:
[tex]\[ \frac{6}{56} = \frac{3}{28} \][/tex]

### Step 4: Compare with Given Options
Now, we compare our result with the given options:
- A. [tex]\(\frac{6}{56}\)[/tex]
- B. [tex]\(\frac{2}{3}\)[/tex]
- C. [tex]\(\frac{3}{56}\)[/tex]
- D. [tex]\(\frac{2}{56}\)[/tex]

The correct option is:
[tex]\[ \boxed{\frac{6}{56}} \text{ or A} \][/tex]

So, the probability that both you and your friend are chosen is [tex]\(\boxed{\frac{6}{56}}\)[/tex], which simplifies to [tex]\(\frac{3}{28}\)[/tex].