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#1 Re: Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 14:31:09

Thank you so much! I have done several practice problems and feel much better about that part of the final. Now am on to sine/cosine...

#2 Re: Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 11:14:43

A bag contains 4 white marbles, 3 blue marbles, and 7 red marbles. What is the probability of drawing two marbles without replacement that are the same color?

I did this:
4c2 divided by 14c2 = 6/91

3c2 = 3, so 3/91
7c2 = 21, so 21/91

So do I add the numerators, or multiply them? I added, for 30/91.

Is that correct?

I can't tell if I am getting this or totally lost. Thank you again.

#3 Re: Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 11:01:47

Oh help. What's the probability of getting a full house consisting of 3 tens and 2 kings?

I am thinking:
the denominator is definitely 2,598,960.

Is the numerator 4c3*4c2 (3 tens out of four, two kings out of 4?)

Thank you thank you thank you.

#4 Re: Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 08:15:30

Another one - if you can check my answer, that would be great!

In a lottery game, I must match 5 cards (in any order) drawn from a standard deck. What's the probability that I'll win the prize?

So it's a combination because order doesn't matter -- 52c5 combination. My calculator has that function, which is lucky because 51! is too big....I get 2,598,960.

So is it ONE out of 2,598,960? I think it is.

#5 Re: Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 08:04:57

I think I get it! Thank you so much for responding so quickly. I am going through several more review questions now and am getting better at this. Thank you again - what a great resource.;)

#6 Help Me ! » Probability - permutation without repetition, with a twist » 2008-06-08 06:31:39

PounderPleaser
Replies: 12

The problem is this:
I have bag of nine straws, in unit lengths from 1 - 9 (say the smallest is 1", the second is 2", and so on up to 9). If I draw three of them out of a bag, what are the chances that the three I draw will create a triangle?

I believe the formula for the denominator (total possible permutations) would be 9!, since I can't choose any length more than once. But since I'm only choosing 3 straws, not all 9, the total is just 9x8x7, or 534. This seems really high to me, but it seems right after I read the permutations description at http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Am I right so far? I hope so, because I'm not even stuck yet. Here's where I get stuck. Not all of these permutations will create a triangle. To create a triangle, the length of the two shortest straws would have to be greater than the length of the longest, right?

Can someone help me with a formula? Final is tomorrow. Thank you so much.

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