Drawing red and blue balls without knowing how many balls are in the bag
This is a lot like the blue and red balls problem, except we don't know how many balls are in the bag, only the relation of red to blue balls, is this problem possible to solve?
Let's say we have a population $x$ of an unknown size $n$ we know that a fraction $f$, let's say $25%$ or $frac{1}{4}$ of that population share a trait (like being a blue ball).
We now randomly select a sample $y$ of a known size $k$, the expected result is that $ktimes f$ balls in our sample are blue, but what is the probability that a number of balls, $p$, in our sample are blue.
So, lets say that I have a bag with $frac{1}{4}$ blue balls and $frac{3}{4}$ red balls, from that bag I take 40 balls. What are the odds that 15 of those balls are blue?
probability statistics
asked Nov 18 at 16:07
Richard Paul Astley
113
add a comment |
This is a lot like the blue and red balls problem, except we don't know how many balls are in the bag, only the relation of red to blue balls, is this problem possible to solve?
Let's say we have a population $x$ of an unknown size $n$ we know that a fraction $f$, let's say $25%$ or $frac{1}{4}$ of that population share a trait (like being a blue ball).
We now randomly select a sample $y$ of a known size $k$, the expected result is that $ktimes f$ balls in our sample are blue, but what is the probability that a number of balls, $p$, in our sample are blue.
So, lets say that I have a bag with $frac{1}{4}$ blue balls and $frac{3}{4}$ red balls, from that bag I take 40 balls. What are the odds that 15 of those balls are blue?
probability statistics
asked Nov 18 at 16:07
Richard Paul Astley
113
2
Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31
add a comment |
This is a lot like the blue and red balls problem, except we don't know how many balls are in the bag, only the relation of red to blue balls, is this problem possible to solve?
Let's say we have a population $x$ of an unknown size $n$ we know that a fraction $f$, let's say $25%$ or $frac{1}{4}$ of that population share a trait (like being a blue ball).
We now randomly select a sample $y$ of a known size $k$, the expected result is that $ktimes f$ balls in our sample are blue, but what is the probability that a number of balls, $p$, in our sample are blue.
So, lets say that I have a bag with $frac{1}{4}$ blue balls and $frac{3}{4}$ red balls, from that bag I take 40 balls. What are the odds that 15 of those balls are blue?
probability statistics
asked Nov 18 at 16:07
Richard Paul Astley
113
This is a lot like the blue and red balls problem, except we don't know how many balls are in the bag, only the relation of red to blue balls, is this problem possible to solve?
Let's say we have a population $x$ of an unknown size $n$ we know that a fraction $f$, let's say $25%$ or $frac{1}{4}$ of that population share a trait (like being a blue ball).
We now randomly select a sample $y$ of a known size $k$, the expected result is that $ktimes f$ balls in our sample are blue, but what is the probability that a number of balls, $p$, in our sample are blue.
So, lets say that I have a bag with $frac{1}{4}$ blue balls and $frac{3}{4}$ red balls, from that bag I take 40 balls. What are the odds that 15 of those balls are blue?
probability statistics
probability statistics
asked Nov 18 at 16:07
Richard Paul Astley
113
asked Nov 18 at 16:07
Richard Paul Astley
113
edited Nov 18 at 16:58
asked Nov 18 at 16:07
Richard Paul Astley
113
asked Nov 18 at 16:07
Richard Paul Astley
113
asked Nov 18 at 16:07
Richard Paul Astley
113
113
2
Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31
add a comment |
2
Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31
2
2
Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31
add a comment |
1 Answer
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For any given $x$ you can compute it. You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 choose 15$ for the number of orders. The answer will be tiny. If you are drawing with replacement $x$ doesn't matter. The chance of getting a blue ball stays $2%$, so the chance of $15$ blue balls out of $40$ is $${40 choose 15}frac {49^{25}}{50^{40}}approx 8cdot 10^{-16}$$
answered Nov 18 at 16:33
Ross Millikan
291k23196371
add a comment |
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1 Answer
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oldest
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votes
For any given $x$ you can compute it. You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 choose 15$ for the number of orders. The answer will be tiny. If you are drawing with replacement $x$ doesn't matter. The chance of getting a blue ball stays $2%$, so the chance of $15$ blue balls out of $40$ is $${40 choose 15}frac {49^{25}}{50^{40}}approx 8cdot 10^{-16}$$
answered Nov 18 at 16:33
Ross Millikan
291k23196371
add a comment |
For any given $x$ you can compute it. You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 choose 15$ for the number of orders. The answer will be tiny. If you are drawing with replacement $x$ doesn't matter. The chance of getting a blue ball stays $2%$, so the chance of $15$ blue balls out of $40$ is $${40 choose 15}frac {49^{25}}{50^{40}}approx 8cdot 10^{-16}$$
answered Nov 18 at 16:33
Ross Millikan
291k23196371
add a comment |
For any given $x$ you can compute it. You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 choose 15$ for the number of orders. The answer will be tiny. If you are drawing with replacement $x$ doesn't matter. The chance of getting a blue ball stays $2%$, so the chance of $15$ blue balls out of $40$ is $${40 choose 15}frac {49^{25}}{50^{40}}approx 8cdot 10^{-16}$$
answered Nov 18 at 16:33
Ross Millikan
291k23196371
For any given $x$ you can compute it. You can compute the chance of drawing specifically $15$ blue balls then $25$ red balls, then multiply by $40 choose 15$ for the number of orders. The answer will be tiny. If you are drawing with replacement $x$ doesn't matter. The chance of getting a blue ball stays $2%$, so the chance of $15$ blue balls out of $40$ is $${40 choose 15}frac {49^{25}}{50^{40}}approx 8cdot 10^{-16}$$
answered Nov 18 at 16:33
Ross Millikan
291k23196371
answered Nov 18 at 16:33
Ross Millikan
291k23196371
answered Nov 18 at 16:33
Ross Millikan
291k23196371
answered Nov 18 at 16:33
Ross Millikan
291k23196371
291k23196371
add a comment |
add a comment |
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Are you drawing with replacement? Note: the phrase "probability is that $ktimes f$ balls in our sample are blue" isn't clear. The expected result is $ktimes f$, true and perhaps that is what you intended. Note that, by linearity of expectation, this result holds regards of whether or not you are replacing.
– lulu
Nov 18 at 16:19
You change from $2%$ blue in the second paragraph to $25%$ in the last. Which is it? It makes a huge difference. Please clarify the question.
– Ross Millikan
Nov 18 at 16:31