Probability (combinatorics) for GRE
Hi everyone...
I'm studying for the GRE. I've been focusing on English because I thought my Math isn't bad but..
it turns out I don't know anything when it comes to questions about probability.
I need a lot of help to learn how to set these questions up.
Here's a question from studying last night:
A 7-member committee is formed from a pool of 10 people.
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I'll help by showing how to think about the first problem:
A 7-member committee is formed from a pool of 10 people. What is the number of the possible outcomes?
There's a little trick that will help you figure out this problem. Creating a 7-member committee from a pool of 10 people is exactly the same as creating a 3-member committee from the same pool of 10 people. Why? Because all that you're doing is dividing the group of 10 people into two groups: one with 7 members, and one with 3 members.
Now, think about the first member in the 3-person pool. There are 10 people, so there are 10 ways to choose the first member.
Think about the second member in the 3-person pool. There are 9 people left, so there are 9 ways left to choose the second member.
Think about the third member in the 3-person pool. There are 8 people left, so there are 8 ways left to choose the third member.
The choices are independent, so it looks like there are 10 * 9 * 8 different 3-member committees from a pool of 10 people.
Is that right? Well... not quite.
You see, for most committees, it doesn't matter whether you chose A first, then B, then C... or chose B first, then C, then A. The order doesn't matter.
So, you need to divide by the number of ways you could order a 3-person committee, to get the real answer. I'll leave figuring that question out to you.
I hope this -- and the reference that I gave above -- helps you with the GRE. Good luck!
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so 10*9*8 divided by the number of ways one can organize three people.. 6? 10*9*8/6?
:) I am glad to know this is actually called "combinatorics". It's a great start, and thank you for the website. I'm studying it now!
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And you're welcome.
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How many ways are there to put 10 people in order? Well, you have 10 choices for the first person, then 9 remaining choices for the 2nd person, then 8 for the 3rd and so on until you have one remaining person to go in the 10th place. So, the number of ways are 10*9*8*7*6*5*4*3*2*1, or 10! (ten factorial). In general, the number of ways to order N things is N!.
Since we're taking only the first 7 to be in the committee, we don't care about the order of the remaining (10-7)=3 people. So we divide by the number of ways to order those 3 people, which is 3! or 6.
If the order of members in the committee mattered (e.g. if there was a president, VP, secretary etc., a permutation), we'd be done. Note that 10!/3! = 10*9*8*7*6*5*4, which is equivalent to saying 10 choices for the 1st member, 9 for the 2nd member, and so on until you have 4 choices for the 7th and final member. Ordering 7 members from a set of 10 can be written as (10 P 7), which equals 10!/(10-7)!. In general, ordering R objects from a set of N objects (N P R) is N!/(N-R)!.
But since the order of the members in the committee doesn't matter (a combination), we need to divide by the number of ways to order 7 people, which is 7!. So, we have 10!/[7!*(10-7)!]. An easy way to calculate this is that 10!/7! is 10*9*8, and (10-7)! is 3! which is 3*2=6, which gives you 10*9*8/6, which is what you had in the previous comment. To sum up, we've chosen 7 members from a set of 10, or (10 C 7), which equals 10!/[7!*(10-7)!]. In general, choosing R members from a set of N objects is often written (N C R), which equals N!/[R!*(N-R)!].
For your second question, you'll want the number of ways you can have combos with exactly 1 veggie, exactly 2 veggies, and exactly 3 veggies. (Do you see why?)
As for standard deviation, it's a measure of the average deviation of each data point from the mean. It applies in this case in the same way it would for bell curves and other data.
Hope this helps!
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