Problem Involving probability with combinations

twoteenine

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Jan 8, 2013
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Here is the problem:
The company Ali works for allows her to invest in her
choice of 10 different mutual funds, 6 of which grew by
at least 5% over the last year. Ali randomly selected 4
of the 10 funds in which to invest. What is the
probability that 3 of Ali’s funds grew by 5%?

I believe that it has to do with combinations, not permutations. I started by labeling the different mutual funds abcdef ghij. a-f are atleast 5%, g-j are not. I believe you can assume g-j are as good as 0% growth but I'm not sure on that. I do not know what to do next. What is the process to solve this?
 
There are \(\displaystyle \displaystyle {10 \choose 6}=210\) ways to choose 6 funds from a total of 10 funds.

There are \(\displaystyle \displaystyle {6 \choose 3}=20\) ways to choose 3 of the 6 funds whose growth rate is at least 5%.

There are \(\displaystyle \displaystyle {4 \choose 1}=4\) ways to choose 1 of the 4 funds whose growth rate is less than 5%.

Hence, the probability in question is:

\(\displaystyle \displaystyle P(X)=\frac{{6 \choose 3}\cdot{4 \choose 1}}{{10 \choose 6}}=?\)
 
Here is the problem:
The company Ali works for allows her to invest in her
choice of 10 different mutual funds, 6 of which grew by
at least 5% over the last year. Ali randomly selected 4
of the 10 funds in which to invest. What is the
probability that 3 of Ali’s funds grew by 5%?

I believe that it has to do with combinations, not permutations. I started by labeling the different mutual funds abcdef ghij. a-f are atleast 5%, g-j are not. I believe you can assume g-j are as good as 0% growth but I'm not sure on that. I do not know what to do next. What is the process to solve this?
I suspect that you can solve this once you understand the question.

You are not being asked what the overall return will be? What the returns were for the six that earned at least 5% is irrelevant as are the returns earned by those that earned less than 5%. Do you see that? That is all just extraneous detail. The question merely wants to know what is the probability that at least three funds of the four randomly selected by Ali were in the group that earned at least 5%. It is a variation on the classic "urn" problem about picking differently colored balls out of an urn.

And yes combinations are the way to go.

Can you do the problem now? If not, please explain what is blocking you.
 
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Thank you! I'm not familiar with the notation you used but I can see what it means. I actually just found a document that explains multiple selection pools. Here it is for anyone that sees this and doesn't understand. I hope it's ok to use a link on this forum.
http://www.bmlc.ca/Math12/Principles of Math 12 - Permutations and Combinations Lesson 2.pdf

Here is another problem.
Four points (A, B, C, and D) lie on one straight line, n,
and five points (E, F, G, H, and J) lie on another
straight line, m, that is parallel to n. What is the
probability that three points, selected at random, will
form a triangle?

I need no more than 2 points from one group, so that the last point in the other group makes a triangle. If I understand how to do this problem I do.

9!/(3!*6!) = 84 total possible combinations to pick 3 points from the nine available
5!/(2!*3!) = 10 combinations to pick 2 points from the (E,F,G,H,J) group
4!/(2!*2!) = 6 combinations to pick 2 points from the (A,B,C and D) group

Probability= (10*6)/84 = 15/21

Did I do this second problem correctly?

Edit: In reply to Jeff, yes I understand that it's not asking for the actual returns. What was blocking me was that I didn't understand how to find the # of combinations with multiple selection pools. Thanks!

2nd Edit: I noticed that I had added the combinations (10+6) instead of multiplying. You are supposed to multiply the combinations aren't you?
 
Last edited:
Thank you! I'm not familiar with the notation you used but I can see what it means. I actually just found a document that explains multiple selection pools. Here it is for anyone that sees this and doesn't understand. I hope it's ok to use a link on this forum.
http://www.bmlc.ca/Math12/Principles of Math 12 - Permutations and Combinations Lesson 2.pdf

Here is another problem.
Four points (A, B, C, and D) lie on one straight line, n,
and five points (E, F, G, H, and J) lie on another
straight line, m, that is parallel to n. What is the
probability that three points, selected at random, will
form a triangle?

I need no more than 2 points from one group, so that the last point in the other group makes a triangle. If I understand how to do this problem I do.

9!/(3!*6!) = 84 total possible combinations to pick 3 points from the nine available
5!/(2!*3!) = 10 combinations to pick 2 points from the (E,F,G,H,J) group
4!/(2!*2!) = 6 combinations to pick 2 points from the (A,B,C and D) group

Probability= (10*6)/84 = 15/21

Did I do this second problem correctly?

Edit: In reply to Jeff, yes I understand that it's not asking for the actual returns. What was blocking me was that I didn't understand how to find the # of combinations with multiple selection pools. Thanks! Ahh OK I think you have it. By the way, one sentence in my prior post was irrelevant. I had not reread the problem before posting my response. I have removed the irrelevant sentence.

2nd Edit: I noticed that I had added the combinations (10+6) instead of multiplying. You are supposed to multiply the combinations aren't you?
If you are asking whether you multiply the combinations of 3 from 6 and 1 from 4 in your original problem before dividing by the combinations of 4 from 10, the answer is in the affirmative. It is dangerous, however, to make generalizations. If the question had asked for the probability of at least 3 funds selected with a return of 5%, you would have needed to use multiplication and addition.
 
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