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Probability question

Dr. J 25,031 3,353 February 18, 2016 at 04:03 PM
.... has me stumped.

It's been awhile since I was an expert in this but I can't figure this out.

Let's say I have an experiment where I draw a marble from a bag, note its color, then replace it, 5 times (5 total picks). The bag has 1 white and 4 black marbles. (the chance of picking a white marble each time is 1/5)

Let's say I run this experiment a bazillion times.

What % of outcomes will have 2 white marbles (only)?

Combinations of 2 of 5 gives 10 (out of 32 total) (31%).

I can't get the math to work out no matter what I do. For the 2 picks, I'd initially say the result would be (10/32)*(1/5)^2*(4/5)^3 but that's an impossibly small number that doesn't add up when you do all combinations

If this could be parameterized that would be great (inputs # of draws and the % of white marbles)

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#2
Interesting question.

I think you start out with, as you did, finding the probability of one ordering of marbles: (1/5)^2*(4/5)^3.

That gives you one ordering, but there are 5! orderings, so you need to adjust the probability to allow for any one of those, so you end up with 5! * (1/5)^2*(4/5)^3.

I'm on my phone, so hopefully that amounts to a number between 0 and 1 Smilie

Edit: no that's not quite right. Many of those orderings are equivalent. Need to think about this some more.

Edit ^ 2: It would be combinations, it would be (5 choose 2) * (1/5)^2*(4/5)^3. 5 choose 2 is 10 rather than 10/32, which is I think what threw you off.

Final answer Smilie
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Last edited by quotidian February 18, 2016 at 04:37 PM
#3
Quote from Dr. J View Post :
.... has me stumped.

It's been awhile since I was an expert in this but I can't figure this out.

Let's say I have an experiment where I draw a marble from a bag, note its color, then replace it, 5 times (5 total picks). The bag has 1 white and 4 black marbles. (the chance of picking a white marble each time is 1/5)

Let's say I run this experiment a bazillion times.

What % of outcomes will have 2 white marbles (only)?

Combinations of 2 of 5 gives 10 (out of 32 total) (31%).

I can't get the math to work out no matter what I do. For the 2 picks, I'd initially say the result would be (10/32)*(1/5)^2*(4/5)^3 but that's an impossibly small number that doesn't add up when you do all combinations

If this could be parameterized that would be great (inputs # of draws and the % of white marbles)
I'm a little rusty on probability too and I'm not motivated to dig up my old textbook.

EDIT: SORRY I misread your post.

If you repeat a bazillion times the answer shouldn't change, your experimental data would just converge on the theoretical result.
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Last edited by jkee February 18, 2016 at 06:26 PM
#4
Okay, back in front of a real computer where I can test my answer, I'm pretty confident it's (5 choose 2) * (1/5)^2*(4/5)^3.

That evaluates to about 20%, which seems about right. I also threw together a quick python script to verify it: https://repl.it/BoOM/1
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#5
Quote from quotidian View Post :
Okay, back in front of a real computer where I can test my answer, I'm pretty confident it's (5 choose 2) * (1/5)^2*(4/5)^3.

That evaluates to about 20%, which seems about right. I also threw together a quick python script to verify it: https://repl.it/BoOM/1
Just add a few zeros and it converges closer to 20.48% as it should (it also takes a lot longer to run).

You got it right. I hadn't used repl.it before that's a handy site. There are a number of web based implementations of octave and some android ones you might enjoy.
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#6
I can't give you the answer but my wife says it sounds like a common core problem LOL.
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#7
Yeah I figured it out later last night.

5C2*(1/5)^2*(4/5)^3 where 5C2 is 10
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#8
Sounds racist, why you want to pick white ones all the time?
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#9
Quote from Dr. J View Post :
Yeah I figured it out later last night.

5C2*(1/5)^2*(4/5)^3 where 5C2 is 10
That looks right, had to get in the way back machine there. Was about to go all Eirenfest Chain on it but you were not drawing between 2 piles. Just felt happy I remembered a big word.

Number of ways to pick * possibility of picking what you want * possibility of not. Seems correct as long as you put the marble back before the next pick and each one is independent which appears to be the case.
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#10
Quote from Pedantyc View Post :
Was about to go all Eirenfest Chain on it but you were not drawing between 2 piles. Just felt happy I remembered a big word.
Yeah, but you spelled it wrong it's ehrenfest
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#11
Quote from jkee View Post :
Yeah, but you spelled it wrong it's ehrenfest
He said he remembered it, not knew how to spell the damn thing laugh out loud
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#12
Quote from Foreveryours View Post :
He said he remembered it, not knew how to spell the damn thing laugh out loud
How pedantic Stick Out Tongue
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