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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
???
Mel001,
I think the required probability is
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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....
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.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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 5/7[sup]n-1[/sup] 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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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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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!
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.
In post#5, mathsy does a cute trick where 7 is in the numerator and denominator!
igloo myrtilles fourmis
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I was trying to solve a similar problem the only difference being that there were 11 balls intead of 7.
Mathsy post kinda helped me solve the problem. Gave me one of those ahah moments. Gotta give my best at math this year to see if I can finally fisnish it :S
Sorry but math for me isn't fun guys ahah.
Gary.
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Hi GaryVin;
Welcome to the forum.
Sorry but math for me isn't fun guys ahah.
That is because you have been doing dry, boring school math. Want to laugh? Look at my stuff. Makes me cry but brings buckets of laughter to everyone else.
You mean that problem was not fun? Surely you felt the adrenalin when you solved it.
Do not underestimate those "Ahahs," they are an endangered species in this world and soon will be extinct.
I saw an ahah yesterday. They usually do not come this far west.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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