Balls and urn problem
$begingroup$
From an urn containing six balls, 3 are white and 3 are black ones a person selects at random an even number of balls ( all different ways of drawing an even number of balls are considered equally probable, irrespective of their number) Then the probability that there will be same number of black and white balls among them
I came to the following cases:
Selevting even number of balls with same black and white
(1W, 1B), (1W, 3B), (2W, 2B), (3W, 3B)
So total cases is 4 and the no of favourable outcome is 3 but still this doesn't give the answer
probability
asked Jan 3 at 1:32
ZerixZerix
1498
$endgroup$
add a comment |
$begingroup$
From an urn containing six balls, 3 are white and 3 are black ones a person selects at random an even number of balls ( all different ways of drawing an even number of balls are considered equally probable, irrespective of their number) Then the probability that there will be same number of black and white balls among them
I came to the following cases:
Selevting even number of balls with same black and white
(1W, 1B), (1W, 3B), (2W, 2B), (3W, 3B)
So total cases is 4 and the no of favourable outcome is 3 but still this doesn't give the answer
probability
asked Jan 3 at 1:32
ZerixZerix
1498
$endgroup$
$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49
add a comment |
$begingroup$
From an urn containing six balls, 3 are white and 3 are black ones a person selects at random an even number of balls ( all different ways of drawing an even number of balls are considered equally probable, irrespective of their number) Then the probability that there will be same number of black and white balls among them
I came to the following cases:
Selevting even number of balls with same black and white
(1W, 1B), (1W, 3B), (2W, 2B), (3W, 3B)
So total cases is 4 and the no of favourable outcome is 3 but still this doesn't give the answer
probability
asked Jan 3 at 1:32
ZerixZerix
1498
$endgroup$
From an urn containing six balls, 3 are white and 3 are black ones a person selects at random an even number of balls ( all different ways of drawing an even number of balls are considered equally probable, irrespective of their number) Then the probability that there will be same number of black and white balls among them
I came to the following cases:
Selevting even number of balls with same black and white
(1W, 1B), (1W, 3B), (2W, 2B), (3W, 3B)
So total cases is 4 and the no of favourable outcome is 3 but still this doesn't give the answer
probability
probability
asked Jan 3 at 1:32
ZerixZerix
1498
asked Jan 3 at 1:32
ZerixZerix
1498
edited Jan 3 at 1:53
Zerix
asked Jan 3 at 1:32
ZerixZerix
1498
asked Jan 3 at 1:32
ZerixZerix
1498
asked Jan 3 at 1:32
ZerixZerix
1498
1498
$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49
add a comment |
$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49
$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49
add a comment |
1 Answer
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$begingroup$
Considering the balls themselves to be distinct and considering balls being drawn to be done so simultaneously (and so order in which balls are drawn doesn't matter):
We are told to consider each outcome in the sample space (the ways in which we can draw an even number of balls total) to be equally likely to occur (irrespective of number of balls drawn or other factors). We then first try to determine the size of the sample space.
The number of ways to draw two balls is $binom{6}{2}$. The number of ways to draw four balls is $binom{6}{4}$. Finally the number of ways to draw six balls is $binom{6}{6}$. We have then $binom{6}{2}+binom{6}{4}+binom{6}{6} = 31$ different possible outcomes in the sample space. ($32$ if you consider drawing zero balls a viable option for our scenario)
The number of ways of drawing two balls such that one is white and one is black is $binom{3}{1}cdot binom{3}{1}=9$ (seen by picking which one white ball was selected followed by which black ball was selected). You can come up with similar values for the number of ways in which you draw four balls where two are white and two are black or six balls where three are white and three are black. Adding these together will give you the total number of ways in which you can draw the same number of white balls as black balls. (If you consider drawing zero balls a viable option for our scenario, then you should account for this here as well since drawing zero balls would satisfy our requirement that the number of white balls be the same as the number of black balls.)
Finally, taking the ratio of the number of "good" options over the total number of options (again because we were told that the outcomes in the sample space in this specific scenario are equally likely) will give us the probability of it occurring.
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
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add a comment |
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$begingroup$
Considering the balls themselves to be distinct and considering balls being drawn to be done so simultaneously (and so order in which balls are drawn doesn't matter):
We are told to consider each outcome in the sample space (the ways in which we can draw an even number of balls total) to be equally likely to occur (irrespective of number of balls drawn or other factors). We then first try to determine the size of the sample space.
The number of ways to draw two balls is $binom{6}{2}$. The number of ways to draw four balls is $binom{6}{4}$. Finally the number of ways to draw six balls is $binom{6}{6}$. We have then $binom{6}{2}+binom{6}{4}+binom{6}{6} = 31$ different possible outcomes in the sample space. ($32$ if you consider drawing zero balls a viable option for our scenario)
The number of ways of drawing two balls such that one is white and one is black is $binom{3}{1}cdot binom{3}{1}=9$ (seen by picking which one white ball was selected followed by which black ball was selected). You can come up with similar values for the number of ways in which you draw four balls where two are white and two are black or six balls where three are white and three are black. Adding these together will give you the total number of ways in which you can draw the same number of white balls as black balls. (If you consider drawing zero balls a viable option for our scenario, then you should account for this here as well since drawing zero balls would satisfy our requirement that the number of white balls be the same as the number of black balls.)
Finally, taking the ratio of the number of "good" options over the total number of options (again because we were told that the outcomes in the sample space in this specific scenario are equally likely) will give us the probability of it occurring.
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
$endgroup$
add a comment |
$begingroup$
Considering the balls themselves to be distinct and considering balls being drawn to be done so simultaneously (and so order in which balls are drawn doesn't matter):
We are told to consider each outcome in the sample space (the ways in which we can draw an even number of balls total) to be equally likely to occur (irrespective of number of balls drawn or other factors). We then first try to determine the size of the sample space.
The number of ways to draw two balls is $binom{6}{2}$. The number of ways to draw four balls is $binom{6}{4}$. Finally the number of ways to draw six balls is $binom{6}{6}$. We have then $binom{6}{2}+binom{6}{4}+binom{6}{6} = 31$ different possible outcomes in the sample space. ($32$ if you consider drawing zero balls a viable option for our scenario)
The number of ways of drawing two balls such that one is white and one is black is $binom{3}{1}cdot binom{3}{1}=9$ (seen by picking which one white ball was selected followed by which black ball was selected). You can come up with similar values for the number of ways in which you draw four balls where two are white and two are black or six balls where three are white and three are black. Adding these together will give you the total number of ways in which you can draw the same number of white balls as black balls. (If you consider drawing zero balls a viable option for our scenario, then you should account for this here as well since drawing zero balls would satisfy our requirement that the number of white balls be the same as the number of black balls.)
Finally, taking the ratio of the number of "good" options over the total number of options (again because we were told that the outcomes in the sample space in this specific scenario are equally likely) will give us the probability of it occurring.
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
$endgroup$
add a comment |
$begingroup$
Considering the balls themselves to be distinct and considering balls being drawn to be done so simultaneously (and so order in which balls are drawn doesn't matter):
We are told to consider each outcome in the sample space (the ways in which we can draw an even number of balls total) to be equally likely to occur (irrespective of number of balls drawn or other factors). We then first try to determine the size of the sample space.
The number of ways to draw two balls is $binom{6}{2}$. The number of ways to draw four balls is $binom{6}{4}$. Finally the number of ways to draw six balls is $binom{6}{6}$. We have then $binom{6}{2}+binom{6}{4}+binom{6}{6} = 31$ different possible outcomes in the sample space. ($32$ if you consider drawing zero balls a viable option for our scenario)
The number of ways of drawing two balls such that one is white and one is black is $binom{3}{1}cdot binom{3}{1}=9$ (seen by picking which one white ball was selected followed by which black ball was selected). You can come up with similar values for the number of ways in which you draw four balls where two are white and two are black or six balls where three are white and three are black. Adding these together will give you the total number of ways in which you can draw the same number of white balls as black balls. (If you consider drawing zero balls a viable option for our scenario, then you should account for this here as well since drawing zero balls would satisfy our requirement that the number of white balls be the same as the number of black balls.)
Finally, taking the ratio of the number of "good" options over the total number of options (again because we were told that the outcomes in the sample space in this specific scenario are equally likely) will give us the probability of it occurring.
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
$endgroup$
Considering the balls themselves to be distinct and considering balls being drawn to be done so simultaneously (and so order in which balls are drawn doesn't matter):
We are told to consider each outcome in the sample space (the ways in which we can draw an even number of balls total) to be equally likely to occur (irrespective of number of balls drawn or other factors). We then first try to determine the size of the sample space.
The number of ways to draw two balls is $binom{6}{2}$. The number of ways to draw four balls is $binom{6}{4}$. Finally the number of ways to draw six balls is $binom{6}{6}$. We have then $binom{6}{2}+binom{6}{4}+binom{6}{6} = 31$ different possible outcomes in the sample space. ($32$ if you consider drawing zero balls a viable option for our scenario)
The number of ways of drawing two balls such that one is white and one is black is $binom{3}{1}cdot binom{3}{1}=9$ (seen by picking which one white ball was selected followed by which black ball was selected). You can come up with similar values for the number of ways in which you draw four balls where two are white and two are black or six balls where three are white and three are black. Adding these together will give you the total number of ways in which you can draw the same number of white balls as black balls. (If you consider drawing zero balls a viable option for our scenario, then you should account for this here as well since drawing zero balls would satisfy our requirement that the number of white balls be the same as the number of black balls.)
Finally, taking the ratio of the number of "good" options over the total number of options (again because we were told that the outcomes in the sample space in this specific scenario are equally likely) will give us the probability of it occurring.
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
answered Jan 3 at 1:49
JMoravitzJMoravitz
48.1k33886
48.1k33886
add a comment |
add a comment |
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$begingroup$
I do not see your logic with what little working you showed. I do agree that the balls ought to be considered distinct for this problem. We have the option of choosing to keep order relevant in or calculations or not when we calculate the sample space size and size of the event in question, and it is simplifies calculations considerably if we choose it to not be... so... have you performed a calculation to see how many ways you can draw an even number of balls to see the size of your sample space? What calculation did you perform and what result did you get? (Also can we draw zero balls?)
$endgroup$
– JMoravitz
Jan 3 at 1:38
$begingroup$
I calculated no of outcome with even number of ball picking. Like (1W, 1B) so that's like 2 ball picking which is even. I have written my possible outcome above for this. For total probability, I must admit I think I have written it wrong
$endgroup$
– Zerix
Jan 3 at 1:45
$begingroup$
I took sample space to be $$ frac{binom{6}{3}$$ which is wrong.
$endgroup$
– Zerix
Jan 3 at 1:49