1. ## Difficult probability question

I've gotten this off another forum (it hasn't been solved yet), and thought you guys might enjoy this problem. I don't need it solved as it's not my homework, so look elsewhere if your goal is to assist people.

You have 9 grey balls, 7 blue balls, 5 green balls in a box. Each ball has an exactly even probability of being selected in a random draw. This even probability property remains the same, even after you've drawn the 1st ball (with or without replacement).

What's the probability that, when you draw 5 balls (without replacement), you'll have selected 3 blue balls and 2 green balls?

My current progress is in spoiler tags.

Spoiler:

2. ## Re: Difficult probability question

Originally Posted by derksheng
I've gotten this off another forum (it hasn't been solved yet), and thought you guys might enjoy this problem. I don't need it solved as it's not my homework, so look elsewhere if your goal is to assist people.

You have 9 grey balls, 7 blue balls, 5 green balls in a box. Each ball has an exactly even probability of being selected in a random draw. This even probability property remains the same, even after you've drawn the 1st ball (with or without replacement).

What's the probability that, when you draw 5 balls (without replacement), you'll have selected 3 blue balls and 2 green balls?

My current progress is in spoiler tags.

Spoiler:

Why doesn't this work:

3. ## Re: Difficult probability question

hi Dragan,

Is that correct?

I now think we should be using permutations for the $|\Omega_1|$.

4. ## Re: Difficult probability question

Originally Posted by derksheng

I now think we should be using permutations...$|\Omega_1|$.
I don't think so because the question at hand states that you're sampling without replacement.

5. ## Re: Difficult probability question

Your solution could very well be correct. A friend's simulation is coming up with 2.2% [don't know how long he ran it for], your solution gives 1.7%.

6. ## Re: Difficult probability question

Originally Posted by derksheng
Your solution could very well be correct.
Yes, it's correct.

7. ## The Following User Says Thank You to Dragan For This Useful Post:

derksheng (08-09-2012)

8. ## Re: Difficult probability question

can you help me on this understanding issue .. Consider the sets of balls A:={a1,a2,a3} and B:={b1,b2,b3,b4} and a random drawing without replacement of 3 balls. which is wrong as our sample space cardinality because we can only have OMEGA={(a,a,a),(a,a,b),(a,b,b),(b,b,b),(b,b,a),(b,a,a),(a,b,a)} and |OMEGA| = 7.

So that would mean our usage of is incorrect?

9. ## Re: Difficult probability question

Originally Posted by derksheng
can you help me on this understanding issue .. Consider the sets of balls A:={a1,a2,a3} and B:={b1,b2,b3,b4} and a random drawing without replacement of 3 balls. which is wrong as our sample space cardinality because we can only have OMEGA={(a,a,a),(a,a,b),(a,b,b),(b,b,b),(b,b,a),(b,a,a),(a,b,a)} and |OMEGA| = 7.

So that would mean our usage of is incorrect?
Notice that first, in your listing, you missed (bab), so there are 8 combinations. Another way to think of this is that you have 2 choices for each of 3 items, so there must be 2^3 possibilities.

More importantly, your calculation of assumes that all 7 objects are distinguishable. For this calculation (a1,b1,b2) is a different outcome than (a1,b1,b3). Your "brute force" listing misses this subtlety.

John

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