Marbles in a Bag Probability Question

  • Thread starter ZZ Specs
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  • #1
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Let's pretend I have a bag of 40 marbles. 19 marbles are red, 21 marbles are blue. If I randomly pick 5 marbles out of the bag, what is the probability that 3 of those 5 marbles are red?
 

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  • #2
phinds
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What do YOU think the answer is? What is your reasoning?

On this forum, we try to help folks learn how to solve problems. We don't spoon feed answers.

And by the way, this is a homework type problem so you are supposed to use the homework template. Please read the forum rules.
 
  • #3
Chronos
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Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.
 
  • #4
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Keep in mind that it makes a difference if you are looking for exactly 3 of 5, or at least 3 of 5.
Keep this in mind, it does in deed make a difference!
 
  • #5
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Hello all! Sorry for the late response, hope you're all still around. Let's pretend I want exactly 3 out of 5, since it's not too hard (I think) to then add the probabilities of 3/5 +(or) 4/5 +(or) 5/5.

I can work it out on a tree, for instance:

19/40*(..conditional probabilities..) + 21/40*(conditional probabilities)

with 5 separate columns (if that makes sense) but I know there has to be a better way.

I understand how basic combinations could give us the chances of pulling 3 red marbles in 3 draws (i.e. all red, or all blue) but I get really messed up when I try to think about 3 red marbles out of 5 draws out of 40 marbles.

It's been a while since I did statistics haha, some hints would be nice! :D And sorry about the homework template / improper forum section, it's also been a bit since I've come to PF. No excuse, of course, totally my fault. I can re-post in the homework section if it's an issue.
 
  • #6
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Don't think about conditional probabilities, just count. You know how many ways there are to draw 5 from 40. Now if 3 are red then you must have chosen 3 from 19 and the remaining 2 from 21. See if that helps and post your answer. Simple counting and multiplication, no need to think harder than that.
 
  • #7
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Yes, 5C40 gives 658,008 ways of picking 5 marbles out of 40.

Does that not ignore certain conditions of non-repetition though? Once you draw the first marble, the probabilities of the next draw have already changed.

I believe it is correct that there are 5! = 120 ways of picking 3 red and 2 blue marbles, if that is anywhere near where you are getting at. Thanks for your help so far!
 
  • #8
HallsofIvy
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One way of picking 3 red and 2 blue marble is "RRRBB".

The probability the first marble you pick is red is, of course, 19/40. Now there are 39 marbles left and 18 are red. The probability that the second marble is red is 18/39. Now there are 38 marbles left and 17 are red. The probability that the third marble is red is 17/38. Now there are 37 marbles left and 21 of them are blue. The probability the fourth marble is blue is 21/37. There are now 36 marbles left and 20 of them are blue. The probability the fifth marble is blue is 20/36.

That will give you the probability of three red marbles followed by two blue marbles.

But if you calculate a different order, say RRBRB, you will see that while you get different fractions, the numerators and denominators are the same, just in different orders. That is, the probability of 'three red and two marbles' in any specific order is the same! You only need to multiply by the number of different orders.

But that is NOT 5!. Since all red are the same and all blue marbles are the same, the number of different orders is
[tex]_5C_3= \begin{pmatrix}5 \\ 3\end{pmatrix}= \frac{5!}{3!2!}= \frac{5(4)}{2}= 10[/tex].
 

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