# Thread: Balls in a jar

1. ## Balls in a jar

I wonder if my approach to this is correct-

Say I have a jar with 10 balls, 9 of which are green and 1 that is red. I start taking out a ball at a time until I get the red ball out. What's the probability of taking 6 balls out? (No repetition)

I thought this might be a case of conditional probability? Basically taking 6 balls out would mean the 6th ball has to be red. so 9/10*8/9*7/8*6/7*5/6*1/5=0.1

Is this correct?

Thanks

2. ## Re: Balls in a jar

Did you mean "no replacement" instead of "repetition"? If not what did you mean and would there be replacement?

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fiffette (08-27-2013)

4. ## Re: Balls in a jar

I mean that you take a ball out and you don't put it back in the jar, you keep it out, then take another one out, and slowly empty the jar.

5. ## Re: Balls in a jar

Once again, I am just OK with probability questions, but I believe this would be:

(1/10 * 1/9 * 1/8 * 1/7 * 1/6 * 1/5)

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fiffette (08-27-2013)

7. ## Re: Balls in a jar

Originally Posted by hlsmith
Once again, I am just OK with probability questions, but I believe this would be:

(1/10 * 1/9 * 1/8 * 1/7 * 1/6 * 1/5)
This doesn't look right to me. The answer they have in the original post is what I think it should be.

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fiffette (08-27-2013)

9. ## Re: Balls in a jar

Thanks Dason, after reviewing post #1 I get it the scenario now.

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fiffette (08-27-2013)

11. ## Re: Balls in a jar

I think another important message in this question is that when you assume each permutation is equally likely, then the position of the target will follows a discrete uniform distribution - i.e. the probability of the target in a specific position will be always 1/10 in your question, regardless of the position you specify.

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fiffette (08-27-2013)

13. ## Re: Balls in a jar

Thanks so much for the help.
There's the next part that got me a bit. I am more confused than I care to admit.

Same jar - 10 balls- 1 red, 9 green.
3 balls are taken out without replacement, several times. (we don't know how many times).
*Clarification: Every 3 balls we take out, we return them back into the jar again, mix the jar, take out 3 again and so on.
Now here comes the monstrous question:
What's the probability that the third time in which one of the balls is red, is the sixth time 3 balls were taken out?

I thought this might be a case of binomial distribution, so I'd call a random variable X to be 'the amount of times we took 3 balls out, one of which is red'
X~B(6,p)
I gathered that P is 0.3 because there are 3 ways/orders in which we can draw a red ball out of a group of 3 balls, and the probability to draw a red ball out of the 10 in the jar is 1/10. so 1/10*3 gives 0.3.
Then it's up to putting things in the formula:
P(x=3)= (3 out of 6)*(0.3)^3*(0.7)^3

However I think that if I pick binomial distribution - order doesn't matter, and here clearly I need the third time of the sixth.

Help!

14. ## Re: Balls in a jar

You probably want to use the negative binomial distribution for that one.

Note that you could use the binomial distribution for this problem but you would also have to use other probability methods. You would need to find the probability that 2 out of the first 5 draws contained red and then multiply that by the probability that the sixth draw contained a red. Do it both with the negative binomial and this way to see that they give the same solution.

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fiffette (08-27-2013)

16. ## Re: Balls in a jar

In that case will be negative binomial which essentially require the last trial to be success/failure (actually is a duality with the Binomial distribution)

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fiffette (08-27-2013)

18. ## Re: Balls in a jar

Originally Posted by BGM
I think another important message in this question is that when you assume each permutation is equally likely, then the position of the target will follows a discrete uniform distribution.
I'm not sure I follow. If there will be repetition+every ball has an equal probability to be taken out then it would be a uniform distribution. Is that what you mean?

19. ## Re: Balls in a jar

Ohh, Dason, you're the boss! Completely makes sense with the binomial distribution method.
Only question though - is it correct to say that the probability of getting a red ball is 0.3 as I calculated above?

Sadly we don't deal with negative binomial distribution in this course so I can't/don't know how to use it :\

20. ## Re: Balls in a jar

Originally Posted by fiffette
Ohh, Dason, you're the boss! Completely makes sense with the binomial distribution method.
Only question though - is it correct to say that the probability of getting a red ball is 0.3 as I calculated above?
Looks fine to me: 1 - (9/10)*(8/9)*(7/8)

Sadly we don't deal with negative binomial distribution in this course so I can't/don't know how to use it :\
But the great thing about learning is that you don't just have to learn what is in the course. The negative binomial distribution is useful and not too bad - take a gander at the wikipedia page.

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fiffette (08-27-2013)

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