drawing balls from urn without out repetition and with repetition
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
add a comment |
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
add a comment |
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
8917
add a comment |
add a comment |
1 Answer
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Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
add a comment |
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1 Answer
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Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
add a comment |
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
add a comment |
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
3,370320
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
add a comment |
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
Could you explain further why you multiply by 5.
– Geralt
Nov 29 '18 at 12:36
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
– platty
Nov 29 '18 at 15:11
add a comment |
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