Post Probability and Counting problems here...

Feed · 04-26-2008, 01:44 PM

Okay, as I mentioned in one of the other threads, probability and counting problems tend to screw me up on the Math section(s). I am sure I'm not the only one out there who experiences this screwup. Post any problems you come across while practicing. We could discuss them together.

Note: If you know the answer to the question, please space it out (like 2-3 lines) from the question before posting it. I would like to try the question myself before glancing at the answer. Thanks!

I'll start:
In a batch of 20 eggs, 4 are broken. If 6 of the eggs are chosen at random, what is the probability that at least 2 of the chosen eggs are broken?

How do you this? O.o

voodoo_santa · 04-26-2008, 02:23 PM

im not 100% sure but since it's at least 2 eggs, you find the probability of 1 egg and 0 egg being broken then u subtract that from 1.

0 egg- .8^6 = .262~

1 egg- (.8^5) * (.2^1) * 6C1 (combination) = .393~

1 - .262 - .393 = .345

So theres around a 34.5% chance that at least 2 eggs are broken if 6 eggs are chosen at random. But like i said I dunno if i did it right...but anyways is that even a real sat problem?

googl-er · 04-26-2008, 03:09 PM

Quote:

Originally Posted by voodoo_santa

but anyways is that even a real sat problem?

I hope not.

Feed · 04-26-2008, 11:28 PM

Ahem, not to stress you out or anything, but this is a real SAT problem.
It's in Barron's ... I'm too lazy to go look up the page no.

So, anyone know for sure how to do this?
(Thanks, voodoo_santa

)

javademon · 04-27-2008, 12:09 AM

Voodoo_santa's got the right idea.

The probability that none of the eggs are broken is the number of ways to choose 6 of the unbroken eggs over the number of ways to choose any six eggs.

In other words, it's 16C6 / 20C6, where C represents combinations - you should know how to do these.

The probability that exactly one egg is broken is equal to

16C5 * 4C1 / 20C6

That is, the number of ways to chose 5 unbroken eggs times the number of ways to chose 1 broken egg, over the total number of ways to choose any six eggs.

Add up the two probabilities and subtract from one. You've got a calculator, so this should be fairly straightforward.

Begoner · 04-27-2008, 08:18 AM

This is definitely not a real SAT problem; while you should be familiar with basic probabilities, they will not expect you to do something as complex as this. Barron's is focused on helping you learn the concepts at a more advanced level in order to boost your confidence and performance on easier problems that will appear on the SAT.

Feed · 04-27-2008, 09:09 AM

Okay, I went back to the book, and looked up where it exactly was.
Begoner's right.

It's only for *extended practice*, and not a real problem on one of the practice tests.

I'm sorry guys. Didn't mean to cause trouble or anything.
Also, thank you Voodoo_santa and javademon, I think I sort of got the problem.