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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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You're working out the number of permutations, but you want the number of combinations.
Order matters with permutations, but not with combinations.
Since it doesn't matter what order you pick the straws in (it won't affect whether or not they can make a triangle since they're the same three straws), combinations are what's needed.
The number of ways of picking r objects from n choices is
.In this case, you'd have
possible combinations of straws.I can't think of any analytical way of finding how many will make a triangle, so I just listed the ones that would:
2 3 4
2 4 5
2 5 6
2 6 7
2 7 8
2 8 9
3 4 5
3 4 6
3 5 6
3 5 7
3 6 7
3 6 8
3 7 8
3 7 9
3 8 9
4 5 6
4 5 7
4 5 8
4 6 7
4 6 8
4 6 9
4 7 8
4 7 9
4 8 9
5 6 7
5 6 8
5 6 9
5 7 8
5 7 9
5 8 9
6 7 8
6 7 9
6 8 9
7 8 9
That's 34 sets, which means the probability of picking a set which will make a triangle is 34/84 = 17/42.
Why did the vector cross the road?
It wanted to be normal.
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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.;)
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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.
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That looks right
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Often, there is a very easy way to compute large combinations.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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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.
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That looks right
Why did the vector cross the road?
It wanted to be normal.
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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.
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Yup, that's right as well.
Why did the vector cross the road?
It wanted to be normal.
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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...
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The probability is 1 in 4,000,000 that a single auto trip in the US results in a fatality. Over a lifetime, the average US driver takes 50,000 trips.
What is the probability of a fatal accident over a lifetime?
I am not sure as to what formula to use for this. Can you help me out please?
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The exact answer can be found with the binomial distribution.
In this case, there are 50,000 trials (car trips), each with a 1/4000000 chance of 'success' (fatality). So we use B(50000, 1/4000000).
The probability of there being a crash is the same as the probability that there aren't no crashes.
Using the Binomial formula, the probability of no crashes is
Note that although it was alright in this case, usually it becomes very hard to use the Binomial distribution when the number of trials is so big. However, the Poisson distribution can be used to approximate it. This approximation gets more accurate as the number of trials increases and the probability of success approaches 0, so in this case it would be very accurate indeed.
B(n,p) is approximated by P(np), so in this case that would be P(0.0125).
Again, to find the probability of a crash we calculate the probability of no crash and take that away from 1.
With a Poisson distribution P(λ), the probability of k occurences is
(In fact, in this case the approximation gave the correct answer to 8 decimal places.)
Why did the vector cross the road?
It wanted to be normal.
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