Question

question 9
at a game show, there are 7 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.
how many ways can you and your friend both be chosen?
a. $_{5}c_{1}=5$
b. $_{7}c_{2}=21$
c. $_{7}p_{3}=210$
d. $_{7}p_{2}=42$

question 10
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 $_{8}c_{3}=56$ total ways to choose the 3 contestants.
what is the probability that you and your friend are both chosen?
a. $\frac{2}{56}$
b. $\frac{6}{56}$
c. $\frac{2}{3}$
d. $\frac{3}{56}$

Explanation:

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Question 9

Step1: Fix you and friend as chosen

We already have 2 of the 3 spots filled (you and your friend), so we only need to choose 1 more person from the remaining 5 people.

Step2: Calculate combinations for 1 spot

The number of ways to choose 1 person from 5 is given by the combination formula $_{n}C_{k}=\frac{n!}{k!(n-k)!}$, where $n=5$, $k=1$.
$$_{5}C_{1}=\frac{5!}{1!(5-1)!}=\frac{5!}{1!4!}=5$$

Step1: Find favorable outcomes

If you and your friend are both chosen, we need 1 more person from the remaining 6 people. The number of favorable ways is $_{6}C_{1}=\frac{6!}{1!(6-1)!}=6$.

Step2: Calculate probability

Probability is the ratio of favorable outcomes to total outcomes. Total outcomes are given as $_{8}C_{3}=56$.
$$\text{Probability}=\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}=\frac{6}{56}$$

Answer:

A. $_{5}C_{1}=5$

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Question 10