Binomial distribution: what is the probability of getting exactly 3 women from a draw of Y = 1 to 10?












0












$begingroup$


So my question is We choose a certain number Y of different people




  1. what is the probability of getting exactly 3 women from a draw of Y = 1 to 10 ?


  2. what is the probability of getting at least more women than men from a draw of Y = 1 to 10



Consider the following: there are 45 people, 38 men and 7 women




  1. I tried for example when Y = 1; P(3 women) = 1C3 * (7/45)^3 * (38/45)^-2


edit: Y = 3 would work in my eyes, and everything above Y = 3 until Y = 10, but under Y = 3 wouldnt be possible in my opinion, or am i wrong there?



my question now is, would that be correct? I have never calculated something like this with a negative exponent. Consider that 1C3 doesnt work..




  1. I tried for example Y = 10; 1 - P( most more women than men) = 1 - 10C3 * (7/45)^7 * (38/45)^3 , but that doesnt look correct in my opinion..






probability combinations negative-binomial







share|cite|improve this question











$endgroup$












  • $begingroup$
    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:27
















0












$begingroup$


So my question is We choose a certain number Y of different people




  1. what is the probability of getting exactly 3 women from a draw of Y = 1 to 10 ?


  2. what is the probability of getting at least more women than men from a draw of Y = 1 to 10



Consider the following: there are 45 people, 38 men and 7 women




  1. I tried for example when Y = 1; P(3 women) = 1C3 * (7/45)^3 * (38/45)^-2


edit: Y = 3 would work in my eyes, and everything above Y = 3 until Y = 10, but under Y = 3 wouldnt be possible in my opinion, or am i wrong there?



my question now is, would that be correct? I have never calculated something like this with a negative exponent. Consider that 1C3 doesnt work..




  1. I tried for example Y = 10; 1 - P( most more women than men) = 1 - 10C3 * (7/45)^7 * (38/45)^3 , but that doesnt look correct in my opinion..






probability combinations negative-binomial







share|cite|improve this question











$endgroup$












  • $begingroup$
    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:27














0












0








0





$begingroup$


So my question is We choose a certain number Y of different people




  1. what is the probability of getting exactly 3 women from a draw of Y = 1 to 10 ?


  2. what is the probability of getting at least more women than men from a draw of Y = 1 to 10



Consider the following: there are 45 people, 38 men and 7 women




  1. I tried for example when Y = 1; P(3 women) = 1C3 * (7/45)^3 * (38/45)^-2


edit: Y = 3 would work in my eyes, and everything above Y = 3 until Y = 10, but under Y = 3 wouldnt be possible in my opinion, or am i wrong there?



my question now is, would that be correct? I have never calculated something like this with a negative exponent. Consider that 1C3 doesnt work..




  1. I tried for example Y = 10; 1 - P( most more women than men) = 1 - 10C3 * (7/45)^7 * (38/45)^3 , but that doesnt look correct in my opinion..






probability combinations negative-binomial







share|cite|improve this question











$endgroup$




So my question is We choose a certain number Y of different people




  1. what is the probability of getting exactly 3 women from a draw of Y = 1 to 10 ?


  2. what is the probability of getting at least more women than men from a draw of Y = 1 to 10



Consider the following: there are 45 people, 38 men and 7 women




  1. I tried for example when Y = 1; P(3 women) = 1C3 * (7/45)^3 * (38/45)^-2


edit: Y = 3 would work in my eyes, and everything above Y = 3 until Y = 10, but under Y = 3 wouldnt be possible in my opinion, or am i wrong there?



my question now is, would that be correct? I have never calculated something like this with a negative exponent. Consider that 1C3 doesnt work..




  1. I tried for example Y = 10; 1 - P( most more women than men) = 1 - 10C3 * (7/45)^7 * (38/45)^3 , but that doesnt look correct in my opinion..






probability combinations negative-binomial




probability combinations negative-binomial






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 17:39









Easymode44

365112




365112










asked Dec 11 '18 at 15:20









LarryLarry

104




104












  • $begingroup$
    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:27


















  • $begingroup$
    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:27
















$begingroup$
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
$endgroup$
– Felix Marin
Dec 11 '18 at 16:27




$begingroup$
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
$endgroup$
– Felix Marin
Dec 11 '18 at 16:27










1 Answer
1






active

oldest

votes


















0












$begingroup$

I think certain things should be cleared up before answering.



If the extraction happens with replacement (which I assume), then the binomial distribution expresses the probability of having $k$ successes in $Y$ trials.



In question $1$, if success is the event "extracting a woman" (happening with probability $p$) then the probability of the event in the question is



$${Ychoose{k}}p^k(1-p)^{Y-k} = {10choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^7$$



Why? Because there are ${10choose{3}}$ possible ways of extracting exactly 3 women and 7 men out of 10 draw (this is the crucial point, the draws are 10!), and the probability of said event does not change.



As for problem 2, try considering all the possible ways in which the event "more women than men have been extracted" can happen. Note that the events "6 women and 4 men", "7 women and 3 men" ecc... are all mutually disjoint, and so you can apply the law of total probability easily.



Addendum



I strongly suspect that the question being asked in point 1 and 2 is asking what the probability of said events is out of 10 draws. However, let's suppose the question actually refers to all possible number of draws from 1 to 10.



If you look carefully at the formula I have written above, it states the probability of having $k$ successes in $Y$ trials.



Suppose we are dealing with $Y=4$. In what ways can I have more women than men in this number of draws? Well, I could have 0 men and 4 women or 1 man and 3 women. These two events happen with respective probability



$${4choose{4}}left(frac{7}{45}right)^4left(frac{38}{45}right)^0$$



$${4choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^1$$



These events are disjoint for any given experiment, so the probability of their union is the sum of their probabilities. Adding them up, we get the desired result.



For each number of draws, you have to make the same considerations. This looks like a rather daunting and tedious exercise, so I would suggest checking if the question was rather not




What is the probability of getting at least more women than men from 10 draws?







share|cite|improve this answer











$endgroup$













  • $begingroup$
    So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
    $endgroup$
    – Larry
    Dec 11 '18 at 16:11










  • $begingroup$
    Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
    $endgroup$
    – Easymode44
    Dec 11 '18 at 16:16










  • $begingroup$
    as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:24










  • $begingroup$
    (for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:26








  • 1




    $begingroup$
    ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
    $endgroup$
    – Larry
    Dec 11 '18 at 16:42













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035395%2fbinomial-distribution-what-is-the-probability-of-getting-exactly-3-women-from-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

I think certain things should be cleared up before answering.



If the extraction happens with replacement (which I assume), then the binomial distribution expresses the probability of having $k$ successes in $Y$ trials.



In question $1$, if success is the event "extracting a woman" (happening with probability $p$) then the probability of the event in the question is



$${Ychoose{k}}p^k(1-p)^{Y-k} = {10choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^7$$



Why? Because there are ${10choose{3}}$ possible ways of extracting exactly 3 women and 7 men out of 10 draw (this is the crucial point, the draws are 10!), and the probability of said event does not change.



As for problem 2, try considering all the possible ways in which the event "more women than men have been extracted" can happen. Note that the events "6 women and 4 men", "7 women and 3 men" ecc... are all mutually disjoint, and so you can apply the law of total probability easily.



Addendum



I strongly suspect that the question being asked in point 1 and 2 is asking what the probability of said events is out of 10 draws. However, let's suppose the question actually refers to all possible number of draws from 1 to 10.



If you look carefully at the formula I have written above, it states the probability of having $k$ successes in $Y$ trials.



Suppose we are dealing with $Y=4$. In what ways can I have more women than men in this number of draws? Well, I could have 0 men and 4 women or 1 man and 3 women. These two events happen with respective probability



$${4choose{4}}left(frac{7}{45}right)^4left(frac{38}{45}right)^0$$



$${4choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^1$$



These events are disjoint for any given experiment, so the probability of their union is the sum of their probabilities. Adding them up, we get the desired result.



For each number of draws, you have to make the same considerations. This looks like a rather daunting and tedious exercise, so I would suggest checking if the question was rather not




What is the probability of getting at least more women than men from 10 draws?







share|cite|improve this answer











$endgroup$













  • $begingroup$
    So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
    $endgroup$
    – Larry
    Dec 11 '18 at 16:11










  • $begingroup$
    Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
    $endgroup$
    – Easymode44
    Dec 11 '18 at 16:16










  • $begingroup$
    as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:24










  • $begingroup$
    (for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:26








  • 1




    $begingroup$
    ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
    $endgroup$
    – Larry
    Dec 11 '18 at 16:42


















0












$begingroup$

I think certain things should be cleared up before answering.



If the extraction happens with replacement (which I assume), then the binomial distribution expresses the probability of having $k$ successes in $Y$ trials.



In question $1$, if success is the event "extracting a woman" (happening with probability $p$) then the probability of the event in the question is



$${Ychoose{k}}p^k(1-p)^{Y-k} = {10choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^7$$



Why? Because there are ${10choose{3}}$ possible ways of extracting exactly 3 women and 7 men out of 10 draw (this is the crucial point, the draws are 10!), and the probability of said event does not change.



As for problem 2, try considering all the possible ways in which the event "more women than men have been extracted" can happen. Note that the events "6 women and 4 men", "7 women and 3 men" ecc... are all mutually disjoint, and so you can apply the law of total probability easily.



Addendum



I strongly suspect that the question being asked in point 1 and 2 is asking what the probability of said events is out of 10 draws. However, let's suppose the question actually refers to all possible number of draws from 1 to 10.



If you look carefully at the formula I have written above, it states the probability of having $k$ successes in $Y$ trials.



Suppose we are dealing with $Y=4$. In what ways can I have more women than men in this number of draws? Well, I could have 0 men and 4 women or 1 man and 3 women. These two events happen with respective probability



$${4choose{4}}left(frac{7}{45}right)^4left(frac{38}{45}right)^0$$



$${4choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^1$$



These events are disjoint for any given experiment, so the probability of their union is the sum of their probabilities. Adding them up, we get the desired result.



For each number of draws, you have to make the same considerations. This looks like a rather daunting and tedious exercise, so I would suggest checking if the question was rather not




What is the probability of getting at least more women than men from 10 draws?







share|cite|improve this answer











$endgroup$













  • $begingroup$
    So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
    $endgroup$
    – Larry
    Dec 11 '18 at 16:11










  • $begingroup$
    Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
    $endgroup$
    – Easymode44
    Dec 11 '18 at 16:16










  • $begingroup$
    as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:24










  • $begingroup$
    (for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:26








  • 1




    $begingroup$
    ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
    $endgroup$
    – Larry
    Dec 11 '18 at 16:42
















0












0








0





$begingroup$

I think certain things should be cleared up before answering.



If the extraction happens with replacement (which I assume), then the binomial distribution expresses the probability of having $k$ successes in $Y$ trials.



In question $1$, if success is the event "extracting a woman" (happening with probability $p$) then the probability of the event in the question is



$${Ychoose{k}}p^k(1-p)^{Y-k} = {10choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^7$$



Why? Because there are ${10choose{3}}$ possible ways of extracting exactly 3 women and 7 men out of 10 draw (this is the crucial point, the draws are 10!), and the probability of said event does not change.



As for problem 2, try considering all the possible ways in which the event "more women than men have been extracted" can happen. Note that the events "6 women and 4 men", "7 women and 3 men" ecc... are all mutually disjoint, and so you can apply the law of total probability easily.



Addendum



I strongly suspect that the question being asked in point 1 and 2 is asking what the probability of said events is out of 10 draws. However, let's suppose the question actually refers to all possible number of draws from 1 to 10.



If you look carefully at the formula I have written above, it states the probability of having $k$ successes in $Y$ trials.



Suppose we are dealing with $Y=4$. In what ways can I have more women than men in this number of draws? Well, I could have 0 men and 4 women or 1 man and 3 women. These two events happen with respective probability



$${4choose{4}}left(frac{7}{45}right)^4left(frac{38}{45}right)^0$$



$${4choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^1$$



These events are disjoint for any given experiment, so the probability of their union is the sum of their probabilities. Adding them up, we get the desired result.



For each number of draws, you have to make the same considerations. This looks like a rather daunting and tedious exercise, so I would suggest checking if the question was rather not




What is the probability of getting at least more women than men from 10 draws?







share|cite|improve this answer











$endgroup$



I think certain things should be cleared up before answering.



If the extraction happens with replacement (which I assume), then the binomial distribution expresses the probability of having $k$ successes in $Y$ trials.



In question $1$, if success is the event "extracting a woman" (happening with probability $p$) then the probability of the event in the question is



$${Ychoose{k}}p^k(1-p)^{Y-k} = {10choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^7$$



Why? Because there are ${10choose{3}}$ possible ways of extracting exactly 3 women and 7 men out of 10 draw (this is the crucial point, the draws are 10!), and the probability of said event does not change.



As for problem 2, try considering all the possible ways in which the event "more women than men have been extracted" can happen. Note that the events "6 women and 4 men", "7 women and 3 men" ecc... are all mutually disjoint, and so you can apply the law of total probability easily.



Addendum



I strongly suspect that the question being asked in point 1 and 2 is asking what the probability of said events is out of 10 draws. However, let's suppose the question actually refers to all possible number of draws from 1 to 10.



If you look carefully at the formula I have written above, it states the probability of having $k$ successes in $Y$ trials.



Suppose we are dealing with $Y=4$. In what ways can I have more women than men in this number of draws? Well, I could have 0 men and 4 women or 1 man and 3 women. These two events happen with respective probability



$${4choose{4}}left(frac{7}{45}right)^4left(frac{38}{45}right)^0$$



$${4choose{3}}left(frac{7}{45}right)^3left(frac{38}{45}right)^1$$



These events are disjoint for any given experiment, so the probability of their union is the sum of their probabilities. Adding them up, we get the desired result.



For each number of draws, you have to make the same considerations. This looks like a rather daunting and tedious exercise, so I would suggest checking if the question was rather not




What is the probability of getting at least more women than men from 10 draws?








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 16:38

























answered Dec 11 '18 at 16:02









Easymode44Easymode44

365112




365112












  • $begingroup$
    So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
    $endgroup$
    – Larry
    Dec 11 '18 at 16:11










  • $begingroup$
    Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
    $endgroup$
    – Easymode44
    Dec 11 '18 at 16:16










  • $begingroup$
    as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:24










  • $begingroup$
    (for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:26








  • 1




    $begingroup$
    ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
    $endgroup$
    – Larry
    Dec 11 '18 at 16:42




















  • $begingroup$
    So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
    $endgroup$
    – Larry
    Dec 11 '18 at 16:11










  • $begingroup$
    Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
    $endgroup$
    – Easymode44
    Dec 11 '18 at 16:16










  • $begingroup$
    as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:24










  • $begingroup$
    (for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
    $endgroup$
    – Larry
    Dec 11 '18 at 16:26








  • 1




    $begingroup$
    ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
    $endgroup$
    – Larry
    Dec 11 '18 at 16:42


















$begingroup$
So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
$endgroup$
– Larry
Dec 11 '18 at 16:11




$begingroup$
So that would be 10C3 * (7/45)^3 * (38/45)^7, which would be equal to = 0.1383042114 = 13.83 % , right? but this works only when Y = 10, what if Y = 1? that is what i dont understand yet...
$endgroup$
– Larry
Dec 11 '18 at 16:11












$begingroup$
Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
$endgroup$
– Easymode44
Dec 11 '18 at 16:16




$begingroup$
Yes, I believe 0.1318... would be right. If you have only one draw (if this is what you mean when you say $Y=1$), then the probability of extracting exactly 3 women is surely 0 (there is not way of extracting 3 women if you have only 1 draw).
$endgroup$
– Easymode44
Dec 11 '18 at 16:16












$begingroup$
as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
$endgroup$
– Larry
Dec 11 '18 at 16:24




$begingroup$
as for problem 2, I would only need to do for example 1C1 * ( 7/45)^1 * (38/45)^0 + 2C2 * ( 7/45)^2 * (38/45)^0 + 3C3 * ( 7/45)^3 * (38/45)^0 + 4C4 * ( 7/45)^4 * (38/45)^0 + 5C5 * ( 7/45)^5 * (38/45)^0 +6C6 * ( 7/45)^6 * (38/45)^0 + etc... , right?
$endgroup$
– Larry
Dec 11 '18 at 16:24












$begingroup$
(for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
$endgroup$
– Larry
Dec 11 '18 at 16:26






$begingroup$
(for problem 1) yes exactly, if Y = 1 then you cannot draw 3 women, for Y = 1 and Y = 2 the result is 0, for Y = 3 it is 13.83% and for the rest for example Y = 6 it would be 6C3 * (7/45)^3 * (38/45)^3, am i right?
$endgroup$
– Larry
Dec 11 '18 at 16:26






1




1




$begingroup$
ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
$endgroup$
– Larry
Dec 11 '18 at 16:42






$begingroup$
ah i see, so this will be a big one... unfortunately, the question is as already mentioned, here i will copy& paste the entire question here: We choose a certain number Y of different people 1)- calculate the probability of getting exactly 3 women from a draw of Y = 1 to 10 2)- calculate the probability of getting at least more women than men from a draw of Y = 1 to 10
$endgroup$
– Larry
Dec 11 '18 at 16:42




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035395%2fbinomial-distribution-what-is-the-probability-of-getting-exactly-3-women-from-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more