What is the probability that after removing all red balls, you have exactly 1 blue and at least 1 green ball...












2












$begingroup$


You have a box with 10 red, 20 blue, and 30 green balls. You take out balls from the box, without returning them.



What is the probability that after you've taken out the last red ball there is exactly one blue ball and at least one green ball remaining?






probability conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
    $endgroup$
    – David G. Stork
    Jan 16 at 22:51






  • 1




    $begingroup$
    The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
    $endgroup$
    – yuner.bekir
    Jan 16 at 22:59










  • $begingroup$
    @yumer.bekir: Do you agree with my rewording of your problem?
    $endgroup$
    – David G. Stork
    Jan 16 at 23:26












  • $begingroup$
    yup. it made the problem easier to understand
    $endgroup$
    – yuner.bekir
    Jan 17 at 11:46
















2












$begingroup$


You have a box with 10 red, 20 blue, and 30 green balls. You take out balls from the box, without returning them.



What is the probability that after you've taken out the last red ball there is exactly one blue ball and at least one green ball remaining?






probability conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
    $endgroup$
    – David G. Stork
    Jan 16 at 22:51






  • 1




    $begingroup$
    The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
    $endgroup$
    – yuner.bekir
    Jan 16 at 22:59










  • $begingroup$
    @yumer.bekir: Do you agree with my rewording of your problem?
    $endgroup$
    – David G. Stork
    Jan 16 at 23:26












  • $begingroup$
    yup. it made the problem easier to understand
    $endgroup$
    – yuner.bekir
    Jan 17 at 11:46














2












2








2





$begingroup$


You have a box with 10 red, 20 blue, and 30 green balls. You take out balls from the box, without returning them.



What is the probability that after you've taken out the last red ball there is exactly one blue ball and at least one green ball remaining?






probability conditional-probability







share|cite|improve this question











$endgroup$




You have a box with 10 red, 20 blue, and 30 green balls. You take out balls from the box, without returning them.



What is the probability that after you've taken out the last red ball there is exactly one blue ball and at least one green ball remaining?






probability conditional-probability




probability conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 0:04









David G. Stork

12.1k41836




12.1k41836










asked Jan 16 at 22:46









yuner.bekiryuner.bekir

204




204












  • $begingroup$
    Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
    $endgroup$
    – David G. Stork
    Jan 16 at 22:51






  • 1




    $begingroup$
    The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
    $endgroup$
    – yuner.bekir
    Jan 16 at 22:59










  • $begingroup$
    @yumer.bekir: Do you agree with my rewording of your problem?
    $endgroup$
    – David G. Stork
    Jan 16 at 23:26












  • $begingroup$
    yup. it made the problem easier to understand
    $endgroup$
    – yuner.bekir
    Jan 17 at 11:46


















  • $begingroup$
    Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
    $endgroup$
    – David G. Stork
    Jan 16 at 22:51






  • 1




    $begingroup$
    The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
    $endgroup$
    – yuner.bekir
    Jan 16 at 22:59










  • $begingroup$
    @yumer.bekir: Do you agree with my rewording of your problem?
    $endgroup$
    – David G. Stork
    Jan 16 at 23:26












  • $begingroup$
    yup. it made the problem easier to understand
    $endgroup$
    – yuner.bekir
    Jan 17 at 11:46
















$begingroup$
Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
$endgroup$
– David G. Stork
Jan 16 at 22:51




$begingroup$
Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully.
$endgroup$
– David G. Stork
Jan 16 at 22:51




1




1




$begingroup$
The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
$endgroup$
– yuner.bekir
Jan 16 at 22:59




$begingroup$
The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur.
$endgroup$
– yuner.bekir
Jan 16 at 22:59












$begingroup$
@yumer.bekir: Do you agree with my rewording of your problem?
$endgroup$
– David G. Stork
Jan 16 at 23:26






$begingroup$
@yumer.bekir: Do you agree with my rewording of your problem?
$endgroup$
– David G. Stork
Jan 16 at 23:26














$begingroup$
yup. it made the problem easier to understand
$endgroup$
– yuner.bekir
Jan 17 at 11:46




$begingroup$
yup. it made the problem easier to understand
$endgroup$
– yuner.bekir
Jan 17 at 11:46










1 Answer
1






active

oldest

votes


















3












$begingroup$

First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $10$ red balls and $20$ blue balls.



If there is exactly one blue ball left, that means that if we take out all $30$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $frac{20}{30} cdot frac{10}{29} = frac{20}{87}$.



From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $60$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $frac{20}{60} cdot frac{10}{59} = frac{10}{177}.$



So the overall probability we want is $frac{20}{87} - frac{10}{177} = frac{890}{5133}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
    $endgroup$
    – yuner.bekir
    Jan 19 at 21:37










  • $begingroup$
    Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
    $endgroup$
    – Misha Lavrov
    Jan 19 at 23:57












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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $10$ red balls and $20$ blue balls.



If there is exactly one blue ball left, that means that if we take out all $30$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $frac{20}{30} cdot frac{10}{29} = frac{20}{87}$.



From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $60$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $frac{20}{60} cdot frac{10}{59} = frac{10}{177}.$



So the overall probability we want is $frac{20}{87} - frac{10}{177} = frac{890}{5133}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
    $endgroup$
    – yuner.bekir
    Jan 19 at 21:37










  • $begingroup$
    Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
    $endgroup$
    – Misha Lavrov
    Jan 19 at 23:57
















3












$begingroup$

First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $10$ red balls and $20$ blue balls.



If there is exactly one blue ball left, that means that if we take out all $30$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $frac{20}{30} cdot frac{10}{29} = frac{20}{87}$.



From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $60$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $frac{20}{60} cdot frac{10}{59} = frac{10}{177}.$



So the overall probability we want is $frac{20}{87} - frac{10}{177} = frac{890}{5133}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
    $endgroup$
    – yuner.bekir
    Jan 19 at 21:37










  • $begingroup$
    Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
    $endgroup$
    – Misha Lavrov
    Jan 19 at 23:57














3












3








3





$begingroup$

First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $10$ red balls and $20$ blue balls.



If there is exactly one blue ball left, that means that if we take out all $30$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $frac{20}{30} cdot frac{10}{29} = frac{20}{87}$.



From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $60$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $frac{20}{60} cdot frac{10}{59} = frac{10}{177}.$



So the overall probability we want is $frac{20}{87} - frac{10}{177} = frac{890}{5133}$.






share|cite|improve this answer









$endgroup$



First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $10$ red balls and $20$ blue balls.



If there is exactly one blue ball left, that means that if we take out all $30$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $frac{20}{30} cdot frac{10}{29} = frac{20}{87}$.



From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $60$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $frac{20}{60} cdot frac{10}{59} = frac{10}{177}.$



So the overall probability we want is $frac{20}{87} - frac{10}{177} = frac{890}{5133}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 1:06









Misha LavrovMisha Lavrov

49k757107




49k757107












  • $begingroup$
    How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
    $endgroup$
    – yuner.bekir
    Jan 19 at 21:37










  • $begingroup$
    Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
    $endgroup$
    – Misha Lavrov
    Jan 19 at 23:57


















  • $begingroup$
    How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
    $endgroup$
    – yuner.bekir
    Jan 19 at 21:37










  • $begingroup$
    Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
    $endgroup$
    – Misha Lavrov
    Jan 19 at 23:57
















$begingroup$
How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
$endgroup$
– yuner.bekir
Jan 19 at 21:37




$begingroup$
How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining
$endgroup$
– yuner.bekir
Jan 19 at 21:37












$begingroup$
Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
$endgroup$
– Misha Lavrov
Jan 19 at 23:57




$begingroup$
Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball).
$endgroup$
– Misha Lavrov
Jan 19 at 23:57


















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