Mixing
From an urn containing a white and b black balls, balls are successively drawn without replacement until only...
3
$begingroup$
From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?
This is what I have done.
(https://i.stack.imgur.com/CZH2x.jpg)
But the answer which is given is :
a/(a+b)
I can't possibly make out how it's happening.
probability
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
add a comment |
3
$begingroup$
From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?
This is what I have done.
(https://i.stack.imgur.com/CZH2x.jpg)
But the answer which is given is :
a/(a+b)
I can't possibly make out how it's happening.
probability
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
3
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
1
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45
add a comment |
3
3
3
$begingroup$
From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?
This is what I have done.
(https://i.stack.imgur.com/CZH2x.jpg)
But the answer which is given is :
a/(a+b)
I can't possibly make out how it's happening.
probability
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
$endgroup$
From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?
This is what I have done.
(https://i.stack.imgur.com/CZH2x.jpg)
But the answer which is given is :
a/(a+b)
I can't possibly make out how it's happening.
probability
probability
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
asked Jan 24 at 5:02
Abhishek GhoshAbhishek Ghosh
878
878
3
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
1
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45
add a comment |
3
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
1
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45
3
3
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
1
1
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
4
$begingroup$
You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.
Also, there is the same probability of leaving any ball last.
So the probability is $frac a{a+b}$
answered Jan 24 at 6:03
abc...abc...
3,207738
$endgroup$
add a comment |
3
$begingroup$
Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = frac{frac{(a+b-1)!}{(a-1)!b!}}{frac{(a+b)!}{a!b!}} = frac{(a+b-1)!}{(a-1)!b!}cdot frac{a!b!}{(a+b)!} = frac{a}{a+b}$$
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
$endgroup$
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
4
$begingroup$
You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.
Also, there is the same probability of leaving any ball last.
So the probability is $frac a{a+b}$
answered Jan 24 at 6:03
abc...abc...
3,207738
$endgroup$
add a comment |
4
$begingroup$
You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.
Also, there is the same probability of leaving any ball last.
So the probability is $frac a{a+b}$
answered Jan 24 at 6:03
abc...abc...
3,207738
$endgroup$
add a comment |
4
4
4
$begingroup$
You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.
Also, there is the same probability of leaving any ball last.
So the probability is $frac a{a+b}$
answered Jan 24 at 6:03
abc...abc...
3,207738
$endgroup$
You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.
Also, there is the same probability of leaving any ball last.
So the probability is $frac a{a+b}$
answered Jan 24 at 6:03
abc...abc...
3,207738
answered Jan 24 at 6:03
abc...abc...
3,207738
answered Jan 24 at 6:03
abc...abc...
3,207738
answered Jan 24 at 6:03
abc...abc...
3,207738
3,207738
add a comment |
add a comment |
3
$begingroup$
Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = frac{frac{(a+b-1)!}{(a-1)!b!}}{frac{(a+b)!}{a!b!}} = frac{(a+b-1)!}{(a-1)!b!}cdot frac{a!b!}{(a+b)!} = frac{a}{a+b}$$
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
$endgroup$
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
add a comment |
3
$begingroup$
Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = frac{frac{(a+b-1)!}{(a-1)!b!}}{frac{(a+b)!}{a!b!}} = frac{(a+b-1)!}{(a-1)!b!}cdot frac{a!b!}{(a+b)!} = frac{a}{a+b}$$
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
$endgroup$
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
add a comment |
3
3
3
$begingroup$
Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = frac{frac{(a+b-1)!}{(a-1)!b!}}{frac{(a+b)!}{a!b!}} = frac{(a+b-1)!}{(a-1)!b!}cdot frac{a!b!}{(a+b)!} = frac{a}{a+b}$$
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
$endgroup$
Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = frac{frac{(a+b-1)!}{(a-1)!b!}}{frac{(a+b)!}{a!b!}} = frac{(a+b-1)!}{(a-1)!b!}cdot frac{a!b!}{(a+b)!} = frac{a}{a+b}$$
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
answered Jan 24 at 6:05
Phil HPhil H
4,2582312
4,2582312
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
add a comment |
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:54
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects.
$endgroup$
– Phil H
Jan 24 at 18:56
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition
$endgroup$
– Abhishek Ghosh
Jan 24 at 19:20
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining.
$endgroup$
– Phil H
Jan 24 at 20:04
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
$begingroup$
Now i got it. And i think it shall be a.(a+b-1)!/(a+b)!
$endgroup$
– Abhishek Ghosh
Jan 24 at 20:10
add a comment |
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3
$begingroup$
Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break.
$endgroup$
– saulspatz
Jan 24 at 5:30
1
$begingroup$
Okay I shall take care of it the next time.
$endgroup$
– Abhishek Ghosh
Jan 24 at 15:45