# Thread: Probably easy but help me out.

1. ## Probably easy but help me out.

Greetings.
The probability question I am about to post is probably straightforward as I think it is but I havent taken a math course in years so maybe Im completely wrong

Anyway here it goes:

'Melissa must choose only 3 of her 8 friends from school to invite to her pool party on the weekend. Melissa cannot decide who to invite so she places 8 of her friends' name into a hat and randomly selects 3. What is the probability Melissa's friend Andrea (who is one of the 8) will be chosen? What is the probability Andrea and another of Melissa's friend, Janice, will BOTH have their names drawn from the hat?'

So is the probability that ANDREA will be chosen 1/8, or 1/3, or is it 1/5 (because 8-3 = 5)???

And for the second part of the question would it be 2/5?

ugh, I dont know. I need to go back to grade 6.

2. Originally Posted by kankabono
Greetings.
The probability question I am about to post is probably straightforward as I think it is but I havent taken a math course in years so maybe Im completely wrong

Anyway here it goes:

'Melissa must choose only 3 of her 8 friends from school to invite to her pool party on the weekend. Melissa cannot decide who to invite so she places 8 of her friends' name into a hat and randomly selects 3. What is the probability Melissa's friend Andrea (who is one of the 8) will be chosen? What is the probability Andrea and another of Melissa's friend, Janice, will BOTH have their names drawn from the hat?'

So is the probability that ANDREA will be chosen 1/8, or 1/3, or is it 1/5 (because 8-3 = 5)???

And for the second part of the question would it be 2/5?

ugh, I dont know. I need to go back to grade 6.
If you've been asked to solve this sort of question then a certain mathematical background is assumed. I will post the unsimplified answers and your job is to think about where these answers have come from and to simplify them:

1. $\displaystyle \frac{{1 \choose 1} \cdot {7 \choose 2}}{{8 \choose 3}}$.

2. $\displaystyle \frac{{2 \choose 2} \cdot {6 \choose 1}}{{8 \choose 3}}$.

3. Thanks. you helped