Required Number of white balls so that 1 white ball is chosen after x picks [closed]
$begingroup$
Lets say that we have a bag of balls (N=100).
We pick 15 balls from the bag (without replacement).
How many white balls (X) should there be in the bag so that we pick at least one white ball with a probability of p>0.5?
How can we calculate this?
What i have done so far is started calculated the probabilities:
First Pick: There is a probability of a1=X/N to find a white ball.
Second Pick: There is a probability of a2=(1-a1)*(X/N-1) to find the ball
Last Pick: There is a probability of a15=(1-a1)(1-a2)...(X/N-14)
Then the total probability to find the ball is the sum of all the above, and for the total probability to be >0.5 we just say sum>0.5 and then solve for X.
I am having problems however implementing this in the actual solution.
probability
$endgroup$
closed as off-topic by GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber♦ Nov 30 '18 at 3:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Lets say that we have a bag of balls (N=100).
We pick 15 balls from the bag (without replacement).
How many white balls (X) should there be in the bag so that we pick at least one white ball with a probability of p>0.5?
How can we calculate this?
What i have done so far is started calculated the probabilities:
First Pick: There is a probability of a1=X/N to find a white ball.
Second Pick: There is a probability of a2=(1-a1)*(X/N-1) to find the ball
Last Pick: There is a probability of a15=(1-a1)(1-a2)...(X/N-14)
Then the total probability to find the ball is the sum of all the above, and for the total probability to be >0.5 we just say sum>0.5 and then solve for X.
I am having problems however implementing this in the actual solution.
probability
$endgroup$
closed as off-topic by GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber♦ Nov 30 '18 at 3:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
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– 5xum
Nov 29 '18 at 8:51
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
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– 5xum
Nov 29 '18 at 8:51
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@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
1
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03
add a comment |
$begingroup$
Lets say that we have a bag of balls (N=100).
We pick 15 balls from the bag (without replacement).
How many white balls (X) should there be in the bag so that we pick at least one white ball with a probability of p>0.5?
How can we calculate this?
What i have done so far is started calculated the probabilities:
First Pick: There is a probability of a1=X/N to find a white ball.
Second Pick: There is a probability of a2=(1-a1)*(X/N-1) to find the ball
Last Pick: There is a probability of a15=(1-a1)(1-a2)...(X/N-14)
Then the total probability to find the ball is the sum of all the above, and for the total probability to be >0.5 we just say sum>0.5 and then solve for X.
I am having problems however implementing this in the actual solution.
probability
$endgroup$
Lets say that we have a bag of balls (N=100).
We pick 15 balls from the bag (without replacement).
How many white balls (X) should there be in the bag so that we pick at least one white ball with a probability of p>0.5?
How can we calculate this?
What i have done so far is started calculated the probabilities:
First Pick: There is a probability of a1=X/N to find a white ball.
Second Pick: There is a probability of a2=(1-a1)*(X/N-1) to find the ball
Last Pick: There is a probability of a15=(1-a1)(1-a2)...(X/N-14)
Then the total probability to find the ball is the sum of all the above, and for the total probability to be >0.5 we just say sum>0.5 and then solve for X.
I am having problems however implementing this in the actual solution.
probability
probability
edited Nov 29 '18 at 10:49
N. F. Taussig
44.1k93356
44.1k93356
asked Nov 29 '18 at 8:44
baskon1baskon1
466
466
closed as off-topic by GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber♦ Nov 30 '18 at 3:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber♦ Nov 30 '18 at 3:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, 5xum, N. F. Taussig, drhab, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
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– 5xum
Nov 29 '18 at 8:51
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
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– 5xum
Nov 29 '18 at 8:51
$begingroup$
@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
1
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03
add a comment |
1
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
$begingroup$
@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
1
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03
1
1
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
1
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
$begingroup$
@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
$begingroup$
@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
1
1
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.
86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.
Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.
$endgroup$
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
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MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
add a comment |
$begingroup$
Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.
Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.
$endgroup$
add a comment |
$begingroup$
As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.
86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.
Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.
$endgroup$
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
add a comment |
$begingroup$
With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.
86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.
Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.
$endgroup$
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
add a comment |
$begingroup$
With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.
86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.
Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.
$endgroup$
With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.
86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.
Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.
answered Nov 29 '18 at 9:07
MBorgMBorg
1751114
1751114
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
add a comment |
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
@baskon1 I answered your question as you wrote it. I am not sure if that is the question you intended to write though given the rather simple solution?
$endgroup$
– MBorg
Nov 29 '18 at 9:10
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
$begingroup$
MBorg, I have changed the question a little bit because obviously it was not very well written (sorry about my English). I don't know if it is allowed to change the question but this is what I was looking for.
$endgroup$
– baskon1
Nov 29 '18 at 10:32
add a comment |
$begingroup$
Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.
Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.
$endgroup$
add a comment |
$begingroup$
Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.
Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.
$endgroup$
add a comment |
$begingroup$
Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.
Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.
$endgroup$
Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.
Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.
edited Nov 29 '18 at 16:24
answered Nov 29 '18 at 11:07
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
$begingroup$
As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.
$endgroup$
add a comment |
$begingroup$
As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.
$endgroup$
add a comment |
$begingroup$
As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.
$endgroup$
As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.
answered Nov 29 '18 at 16:33
Michael HoppeMichael Hoppe
11k31836
11k31836
add a comment |
add a comment |
1
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Nov 29 '18 at 8:51
$begingroup$
@5xum I don't see your close vote.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:55
1
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 29 '18 at 10:49
$begingroup$
@5xum OP has done what you've said.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:03