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October 15th, 2013, 03:37 AM   #1
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Probability of Books Taken

There are two boxes: Box A and Box B. Box A contains 5 statistic books and 3 calculus books. Box B contains 4 statistic books and 2 calculus books. A coin is thrown twice. If at least 1 Head appears, a book from box A will be taken. Otherwise, a book from box B will be taken.
a. If 2 books are taken, what is the probability that those are calculus books?
b. P (2 statistic books, 1 calculus books)
c. If 3 statistic books are taken, what is the probability that those books come from box B?
d. If a statistic book and a calculus book are taken, what is the probability that those books come from box A?

I'm doing the a question like this:
(3/4)(3C2)/(8C2) + (1/4)(2C2)/(6C2)
Am I right?

How to do the b question? I have no idea at all.

I'm doing the a question like this:
(1/4)(4C3)/(6C3)
Am I right?

I also have no idea whatsoever to do the d question. How to do it?

Thanks in advance.
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October 15th, 2013, 04:38 AM   #2
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Re: Probability of Books Taken

Question a)

I don't think your answer to a) is correct. The problem is quite complicated, but basically the coin tossing results in 3/4 probability of taking a book from box A and a 1/4 probablity of taking a book from box B.

When it says "two books" are taken, I assume they mean that the coin tossing happens twice, so you have:

(3/4)(3/4) = 9/16 that both books come from box A
2(3/4)(1/4) = 6/16 that one book comes from box A and one from box B
((1/4)(1/4) = 1/16 that both books come from box B

I would redo your calculations based on this.

Also, although you get the right answer using binomials, this seems to me a complicated way to do it. The probability of picking two calculus books from box A (given that both books come from box A) is simply:

(3/(2/7) = 6/56 = 3/28

And two calculus books from box B (given both books come from box B) is simply:

(2/6)(1/5) = 2/20 = 1/15

And one calculus book from A and one from B (given that one book comes from each box) is:

(3/(2/6) = 1/8

You can put this altogether now to get the probability of two calculus books being chosen as:

(Note: this is where I used the conditional probailities calculated above:)

(9/16)(3/2 + (1/16)(1/15) + (6/16)(1/

Question b)

Please explain what question b is. Are you saying that 3 books are picked and 2 are stats and 1 calculus? If so, it is similar to the above, only more complicated.

Questions c and d

These are harder and require Bayes Theorem and/or an understanding of conditional probability. Have you covered Bayes Theorem?
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October 15th, 2013, 06:30 AM   #3
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Re: Probability of Books Taken

Yes. Question b is just as you think. How to do it since we have so many possibilities?

I have covered both Bayes Theorem and conditional probability. However, I still don't quite get it.
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October 15th, 2013, 08:19 AM   #4
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Re: Probability of Books Taken

Quote:
Originally Posted by Monox D. I-Fly
Yes. Question b is just as you think. How to do it since we have so many possibilities?
Well, you just work through them all! I can't see any short-cuts.

Quote:
Originally Posted by Monox D. I-Fly

I have covered both Bayes Theorem and conditional probability. However, I still don't quite get it.
For c). First you calculate the probability of getting 3 stats books (any way) = P(3S), say. Then, or at the same time, you calculate the probability of getting 3 stats books from box B = P(3S-B) - with the coin tossing and everything!.

Then, given that you have ended up with 3 stats books, the probability they came from box B is:

P(3S-B)/P(3S)

The way I think of this is to imagine doing this experiment over and over and counting how many times you get 3 stats books and how many times they all came from box B. Then, if all you know is that you have 3 stats books, you can see how likely it is that they came from box B. All the other outcomes are effectively irrelevant. It's only the relative probability of these two events that matters.

Are you able to calculate P(3S) and P(3S-B)?
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October 16th, 2013, 01:51 AM   #5
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Re: Probability of Books Taken

Quote:
Originally Posted by Pero
Also, although you get the right answer using binomials, this seems to me a complicated way to do it. The probability of picking two calculus books from box A is simply:

(3/(2/7) = 6/56 = 3/28

And two calculus books from box B is simply:

(2/6)(1/5) = 2/20 = 1/15

And one calculus book from A and one from B is:

(3/(2/6) = 1/8

You can put this altogether now to get the probability of two calculus books being chosen as:

(9/16)(3/2 + (1/16)(1/15) + (6/16)(1/
Wait. Now that I think about it, seems you forgot that the probability of a book taken from box A is 3 times bigger than if it is taken from box B instead. Thus, the probability of picking two calculus books from box A becomes 3(3/(2/7) = 18/56 = 9/28. But, how to apply it to the probability of one calculus book from A and one from B?

Quote:
Originally Posted by Pero
Are you able to calculate P(3S) and P(3S-B)?
Ummm, maybe. I'll reread Bayes Theorem and conditional probability later.
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October 16th, 2013, 02:20 AM   #6
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Re: Probability of Books Taken

Quote:
Originally Posted by Monox D. I-Fly

Wait. Now that I think about it, seems you forgot that the probability of a book taken from box A is 3 times bigger than if it is taken from box B instead. Thus, the probability of picking two calculus books from box A becomes 3(3/(2/7) = 18/56 = 9/28. But, how to apply it to the probability of one calculus book from A and one from B?
This is not right. The probability of picking one book from box A is 3 time greater, but the probability of both books coming from box A is 9 times greater than both books coming from box B - but you have the probability of getting one book from each box to consider as well.

That's why I split the calculation up into the probability of how many books came from each box and then, given that, the probability of getting two calculus books. Which is a conditional probability.

I've edited my original solution to make this clear.
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October 16th, 2013, 02:41 AM   #7
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Re: Probability of Books Taken

Quote:
Originally Posted by Pero
That's why I split the calculation up into the probability of how many books came from each box and then, given that, the probability of getting two calculus books. Which is a conditional probability.

I've edited my original solution to make this clear.
OK. I've understood your explanation, but there's still one thing I don't understand:
Quote:
Originally Posted by Pero
2(3/4)(1/4) = 6/16 that one book comes from box A and one from box B
Why did you have to multiply it by two?
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October 16th, 2013, 03:09 AM   #8
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Re: Probability of Books Taken

This is a general point that occurs in many situations. There is only one way to get two books from box A, but there are two ways to get one book from A and one from B. The first book could be from A and the second from B or vice versa. Each of these events has a probability of (3/4)(1/4).

The best way to understand this is to break the problem down (even further). There are four possibilities:

AA with probability (3/4)(3/4)
AB with probability (3/4)(1/4)
BA with probability (1/4)(3/4)
BB with probability (1/4)(1/4)

If you have three books, then there are 3 ways to get one book from A (all with the same probability of (3/4)(1/4)(1/4)) and three ways to get one book from B (al with probability (3/4)(3/4)(1/4).

So, if you try problem b), you will need to see this.

In general, for more books, you have the Binomial Theorem. The above is an example for n = 2 and n = 3.

In general, the probability of getting k books out of n from box A is:



Knowing this is critical to doing these sort of problems.
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October 16th, 2013, 03:14 AM   #9
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Re: Probability of Books Taken

OK,thank you very much for all your help and efforts. Seems like I'm really slow in digesting this. Again, thank you very much.
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