Expected number of the trials
So here is the question:
There is a bag of marbles with an equal number of black and red marbles in it. I randomly draw marbles. I will stop drawing when I have at least one black marble and at least one red marble. What is the expected number of marbles that I draw?
I couldn’t find which probability distribution to use to solve this question. First, I thought that this is a geometric but without replacement. Then, I thought that this is a hypergeometric but i couldn’t find the answer anyway. Help me to solve this question,
Thanks.
probability expected-value
asked Nov 25 at 14:56
Efe Gürtunca
1
|
show 3 more comments
So here is the question:
There is a bag of marbles with an equal number of black and red marbles in it. I randomly draw marbles. I will stop drawing when I have at least one black marble and at least one red marble. What is the expected number of marbles that I draw?
I couldn’t find which probability distribution to use to solve this question. First, I thought that this is a geometric but without replacement. Then, I thought that this is a hypergeometric but i couldn’t find the answer anyway. Help me to solve this question,
Thanks.
probability expected-value
asked Nov 25 at 14:56
Efe Gürtunca
1
Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17
|
show 3 more comments
So here is the question:
There is a bag of marbles with an equal number of black and red marbles in it. I randomly draw marbles. I will stop drawing when I have at least one black marble and at least one red marble. What is the expected number of marbles that I draw?
I couldn’t find which probability distribution to use to solve this question. First, I thought that this is a geometric but without replacement. Then, I thought that this is a hypergeometric but i couldn’t find the answer anyway. Help me to solve this question,
Thanks.
probability expected-value
asked Nov 25 at 14:56
Efe Gürtunca
1
So here is the question:
There is a bag of marbles with an equal number of black and red marbles in it. I randomly draw marbles. I will stop drawing when I have at least one black marble and at least one red marble. What is the expected number of marbles that I draw?
I couldn’t find which probability distribution to use to solve this question. First, I thought that this is a geometric but without replacement. Then, I thought that this is a hypergeometric but i couldn’t find the answer anyway. Help me to solve this question,
Thanks.
probability expected-value
probability expected-value
asked Nov 25 at 14:56
Efe Gürtunca
1
asked Nov 25 at 14:56
Efe Gürtunca
1
asked Nov 25 at 14:56
Efe Gürtunca
1
asked Nov 25 at 14:56
Efe Gürtunca
1
asked Nov 25 at 14:56
Efe Gürtunca
1
1
Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17
|
show 3 more comments
Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17
Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17
|
show 3 more comments
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Are you drawing with or without replacement?
– lulu
Nov 25 at 15:08
Assuming you meant "without replacement", a good way to get started is to work it explicitly for small collections. That should give you some ideas about a general solution and it will give you several solid values that you can test your eventual expression against.
– lulu
Nov 25 at 15:10
I tried to draw without replacement and as you said I found the expected value when there are 4 marbles and 6 marbles.(when 4 marbles expected value=7/3 and when 6 marbles expected value=25/10). Then, I tried to find a general expression for the solution but I couldn't.
– Efe Gürtunca
Nov 25 at 15:14
Well, assuming you start with $N$ of each, what's the probability of drawing $i$ red followed by black?
– lulu
Nov 25 at 15:14
To make it look more like a standard hypergeometric problem, note that the first draw doesn't really matter. It's either red or black, and in either case you are down to a standard hypergeometric calculation. That should simplify things.
– lulu
Nov 25 at 15:17