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I have a bag with 5 pennies and 3 nickels. I draw coins out one at a time at random. What is the probability that I haven't removed all 3 nickels after 4 draws?
I tried 1/2 and 11/14. They are wrong
I see you have graph paper.
You must be plotting something
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Hi;
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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This solution is not right, i tried it. Can you show how you got it, or something
I think he might be right. I am getting 13/14.
(Select[Permutations[{p, p, p, p, p, n, n, n}], Count[Take[#, 4], n] < 3 &] // Length)/(Permutations[{p, p, p, p, p, n, n, n}] // Length)
Last edited by anonimnystefy (2013-09-05 12:20:32)
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Hi all;
That is all the ways to draw 4 coins from 5 and 3. There are 15 total ways and four winners. So I will have to go with 11 / 15 jist as before. Also, the OP already verified this.
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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But, the probabilities of those 4-draws are not all the same.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Here's a simulation that confirms 13/14:
(Table[If[Count[
Delete[Delete[
Delete[Delete[{p, p, p, p, p, n, n, n}, RandomInteger[{1, 8}]],
RandomInteger[{1, 7}]], RandomInteger[{1, 6}]],
RandomInteger[{1, 5}]], n] > 0, 1, 0], {100000}]//Total)/100000//N
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Hooohaa as the great Al Pacino would say. You have provided some good evidence, I will rethink the whole thing as soon as I come back.
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.
Offline
Okay. See you later.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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I think the answer is 13/14.
(5C4*3C0 + 5C3*3C1 + 5C2*3C2)/8C4 = 13/14.
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Hi;
Yes, that is the correct answer. Verified by the using the hypergeometric distribution.
Thank you spotting that everyone.
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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Hello;
Probability[x < 3,
x \[Distributed] HypergeometricDistribution[4, 3, 8]]
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For the other M:
Probability(RandomVariable(Hypergeometric(8, 3, 4)) < 3)
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.
Offline
Maxima:
sum(pdf_hypergeometric(i,3,5,4),i,0,2);
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?
"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
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