Math IIC probabilities





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Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: Math IIC probabilities
By Jared (Jared) on Thursday, June 03, 2004 - 07:37 pm: Edit

The Math IIC is coming up in 2 days and im still terrible at probabilities...can someone please explain to me in detail when youre supposed to use combinations, when youre supposed to use permutations, when youre supposed to use factorials, and when youre just supposed to multiply the numbers together. Thanks a lot!

By Chamonix (Chamonix) on Thursday, June 03, 2004 - 07:44 pm: Edit

I love those ones. Let's see, probabilty...say the question says what's the probablity when taking two cards out of a deck of getting two 10's. This is a generic probability one. The answer is (4/52)*(3/51). Remember to change the base from 52 to 51. As for combinations, I don't know how to teach them. I just enter them into my calculator and it does the work.

By Metra (Metra) on Thursday, June 03, 2004 - 07:45 pm: Edit

Thats easy, it gets a little complicated with "or" and "at least"

By Z00b (Z00b) on Thursday, June 03, 2004 - 08:19 pm: Edit

For or at least, you have to use the negation, I think. For example, what is the probability of getting one head or one six on a die? The probability of getting neither is (1/2) * (5/6) so there is a 5/12 probability of getting neither, which means there is a 7/12 probability of getting at least one of them. Am I doing this right?

By Zantedeschia (Zantedeschia) on Thursday, June 03, 2004 - 09:18 pm: Edit

You also have to subtract the probability of getting both.

By 1212 (1212) on Thursday, June 03, 2004 - 09:37 pm: Edit

theres some confusion in the wording of Z00b's problem. probability of getting at least one of them = 7/12, the probability of getting one head or one six on a die = 1/2

By Scion (Scion) on Thursday, June 03, 2004 - 09:46 pm: Edit

for or, ADD the probabilities

for at least, do (1-probability of none)

...hope that kinda helps!

By Lahlahlah (Lahlahlah) on Thursday, June 03, 2004 - 09:51 pm: Edit

ummm not really the probability of getting one head = 1/2 the probability of getting one six= 1/6 the probability of both = 1/2 + 1/6 - the probability of getting both (1/12) which equals 7/12.

By Scion (Scion) on Friday, June 04, 2004 - 12:44 am: Edit

o yeah, u gotta subtract the probability of both.

By Aim78 (Aim78) on Friday, June 04, 2004 - 01:04 am: Edit

How the hell do you get a head on a die?


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