Difficult probability question

[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:

Construct two sample spaces, and . Let be every possible combination of 5 balls drawn (without replacement) out of the 21 total balls. Let be every possible combination of 5 balls drawn (without replacement) out of the 21 balls that satisfies the constraint of being 3 blue and 2 green balls.

We then use the cardinality: (combination).

Finally, we have to find , and divide this into to arrive at our solution.

Can't do this yet.

[Dragan]

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:

Construct two sample spaces, and . Let be every possible combination of 5 balls drawn (without replacement) out of the 21 total balls. Let be every possible combination of 5 balls drawn (without replacement) out of the 21 balls that satisfies the constraint of being 3 blue and 2 green balls.

We then use the cardinality: (combination).

Finally, we have to find , and divide this into to arrive at our solution.

Can't do this yet.

Why doesn't this work:

[derksheng]

hi Dragan,

Is that correct?

I now think we should be using permutations for the [itex]|\Omega_1|[/itex].

[Dragan]

I now think we should be using permutations...[itex]|\Omega_1|[/itex].

I don't think so because the question at hand states that you're sampling without replacement.

[derksheng]

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%.

[Dragan]

Your solution could very well be correct.

Yes, it's correct.

[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?

[SmoothJohn]

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