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#1 2006-04-16 18:04:31
- Mel001
- Guest
probability problem
Hey everyone,
I am very insecure about the following probability question. Would anyone like to have a look at my answer and tell me whether what I'm doing is right/wrong? Thanks so much!
PROBLEM:
A bag contains 2 red and 5 green balls. Balls are repeatedly drawn from the bag with replacement . Find the probability that a red ball is drawn on the nth draw for the first time.
SOLUTION:
p(red) = 2/7
p(green) = 5/7
probability of being red after n green draws:
Event E: red and n *green
F: n*green
p(E/F) = p(E and F) / p(F)
now:
p(E and F) = p (E) = (5/7)*(2/7)^n
p(E) = (5/7)^n
Thus:
p(E/F) = p(E and F)/p(E) = (2/7)^n*(5/7) /(5/7)^n
???
#3 2006-04-16 19:23:52
- Mel001
- Guest
Re: probability problem
Thanks for that!!! It's much appreciated.
Would you mind tellin me what your thoughts were when finding the answer? And do you possibly see what went wrong in my attempt and if one can tackle the problem with conditional probabilities?
When simplyfying my expression I got:
(5/7)^n
Not sure what I did wrong....
#4 2006-04-16 23:25:05
- ganesh
- Moderator
Re: probability problem
If the probability of an even occuring is n, the probability of it not occuring is 1-n.
I think you were applying the formula for Binomial Distribution.
Character is who you are when no one is looking.
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#5 2006-04-16 23:57:46
- mathsyperson
- Moderator
Re: probability problem
I get a different answer to both of those.
How I see it, you'll have to pick a red ball on the nth time, which there is a 2/7 chance of. Before the nth time, you'll need to always pick green balls so that the nth time is the 1st time you get a red one. That means that you need to pick a green ball (n-1) times, which there is a 5/7n-1 chance of.
Therefore the total probability is:
Or even:
Why did the vector cross the road?
It wanted to be normal.
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#6 2006-04-17 10:18:24
- MathsIsFun
- Administrator
Re: probability problem
I went crazy and made a simulation of this (one day I hope to turn it into a general "marbles simulator").
Anyway, the numbers come out similar to mathsy's formula
You can try it yourself here: Marble Probability Simulation
(Note: "Draws" is "n")
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#7 2006-04-17 16:11:37
- Mel001
- Guest
Re: probability problem
Thanks for your help!
Your answer does makes sense (a lot more than mine). Still haven't figured out what exactly is wrong with my calculation, but I will have another go and let you know, if I find out.
That simulator is quite neat btw!
#8 2008-03-28 09:48:23
- Ross
- Guest
Re: probability problem
This is just the commonly known goemetric distribution,
p(x) = p * (1-p)^n-1
where p is the probability of your event (choosing a red ball on any independent draw).
http://en.wikipedia.org/wiki/Geometric_distribution
So Mathsy is correct.
#9 2008-03-28 12:49:41
- John E. Franklin
- Super Member
Re: probability problem
In post#5, mathsy does a cute trick where 7 is in the numerator and denominator!
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