Mixing
Probability exercise about allocating bottles of wine
$begingroup$
The exercise states the following:
We have 15 bottles of wine that we will randomly distribute among three customers: A will get 2, B will get 8 and C will get 5. We've learned later that 4 of this 15 bottles are corked: what are the chances that customer B doesn't get a single corked bottle of wine?
I'm really having problems facing this kind of problem in which we use permutations and combinations to find the probability of an event. That's why I'd be thankful if someone presents me multiple methods to solve this exercise and explains me the intuition behind each one.
Moreover, I'd appreciate to be explained what kind of probability distribution it follows, because I think it might follow a binomial distribution but I'm not able to present the problem in such terms.
probability combinatorics probability-distributions
edited Jan 20 at 15:44
Hanno
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
$endgroup$
add a comment |
$begingroup$
The exercise states the following:
We have 15 bottles of wine that we will randomly distribute among three customers: A will get 2, B will get 8 and C will get 5. We've learned later that 4 of this 15 bottles are corked: what are the chances that customer B doesn't get a single corked bottle of wine?
I'm really having problems facing this kind of problem in which we use permutations and combinations to find the probability of an event. That's why I'd be thankful if someone presents me multiple methods to solve this exercise and explains me the intuition behind each one.
Moreover, I'd appreciate to be explained what kind of probability distribution it follows, because I think it might follow a binomial distribution but I'm not able to present the problem in such terms.
probability combinatorics probability-distributions
edited Jan 20 at 15:44
Hanno
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
$endgroup$
add a comment |
$begingroup$
The exercise states the following:
We have 15 bottles of wine that we will randomly distribute among three customers: A will get 2, B will get 8 and C will get 5. We've learned later that 4 of this 15 bottles are corked: what are the chances that customer B doesn't get a single corked bottle of wine?
I'm really having problems facing this kind of problem in which we use permutations and combinations to find the probability of an event. That's why I'd be thankful if someone presents me multiple methods to solve this exercise and explains me the intuition behind each one.
Moreover, I'd appreciate to be explained what kind of probability distribution it follows, because I think it might follow a binomial distribution but I'm not able to present the problem in such terms.
probability combinatorics probability-distributions
edited Jan 20 at 15:44
Hanno
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
$endgroup$
The exercise states the following:
We have 15 bottles of wine that we will randomly distribute among three customers: A will get 2, B will get 8 and C will get 5. We've learned later that 4 of this 15 bottles are corked: what are the chances that customer B doesn't get a single corked bottle of wine?
I'm really having problems facing this kind of problem in which we use permutations and combinations to find the probability of an event. That's why I'd be thankful if someone presents me multiple methods to solve this exercise and explains me the intuition behind each one.
Moreover, I'd appreciate to be explained what kind of probability distribution it follows, because I think it might follow a binomial distribution but I'm not able to present the problem in such terms.
probability combinatorics probability-distributions
probability combinatorics probability-distributions
edited Jan 20 at 15:44
Hanno
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
edited Jan 20 at 15:44
Hanno
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
edited Jan 20 at 15:44
Hanno
2,274628
edited Jan 20 at 15:44
Hanno
2,274628
edited Jan 20 at 15:44
Hanno
2,274628
2,274628
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
asked Jan 20 at 15:10
torito verdejotorito verdejo
745
745
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probability distribution involved here is the hypergeometric one.
The probability that customer B did only get bottles of wine that were not corked equals:$$frac{binom{11}{8}binom{4}{0}}{binom{15}{8}}=frac{binom{11}{8}}{binom{15}{8}}$$
For a more direct way to calculate this probability see the answer of Ross.
answered Jan 20 at 15:29
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
For this problem, ignore A and C. You are just drawing $8$ bottles for B without replacement and need them all to be good. What is the chance the first bottle is good? Assuming it is, what is the chance the second is good? Keep going. The chances decrease because you are using up good bottles.
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability distribution involved here is the hypergeometric one.
The probability that customer B did only get bottles of wine that were not corked equals:$$frac{binom{11}{8}binom{4}{0}}{binom{15}{8}}=frac{binom{11}{8}}{binom{15}{8}}$$
For a more direct way to calculate this probability see the answer of Ross.
answered Jan 20 at 15:29
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The probability distribution involved here is the hypergeometric one.
The probability that customer B did only get bottles of wine that were not corked equals:$$frac{binom{11}{8}binom{4}{0}}{binom{15}{8}}=frac{binom{11}{8}}{binom{15}{8}}$$
For a more direct way to calculate this probability see the answer of Ross.
answered Jan 20 at 15:29
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The probability distribution involved here is the hypergeometric one.
The probability that customer B did only get bottles of wine that were not corked equals:$$frac{binom{11}{8}binom{4}{0}}{binom{15}{8}}=frac{binom{11}{8}}{binom{15}{8}}$$
For a more direct way to calculate this probability see the answer of Ross.
answered Jan 20 at 15:29
drhabdrhab
102k545136
$endgroup$
The probability distribution involved here is the hypergeometric one.
The probability that customer B did only get bottles of wine that were not corked equals:$$frac{binom{11}{8}binom{4}{0}}{binom{15}{8}}=frac{binom{11}{8}}{binom{15}{8}}$$
For a more direct way to calculate this probability see the answer of Ross.
answered Jan 20 at 15:29
drhabdrhab
102k545136
answered Jan 20 at 15:29
drhabdrhab
102k545136
answered Jan 20 at 15:29
drhabdrhab
102k545136
answered Jan 20 at 15:29
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
For this problem, ignore A and C. You are just drawing $8$ bottles for B without replacement and need them all to be good. What is the chance the first bottle is good? Assuming it is, what is the chance the second is good? Keep going. The chances decrease because you are using up good bottles.
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
For this problem, ignore A and C. You are just drawing $8$ bottles for B without replacement and need them all to be good. What is the chance the first bottle is good? Assuming it is, what is the chance the second is good? Keep going. The chances decrease because you are using up good bottles.
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
For this problem, ignore A and C. You are just drawing $8$ bottles for B without replacement and need them all to be good. What is the chance the first bottle is good? Assuming it is, what is the chance the second is good? Keep going. The chances decrease because you are using up good bottles.
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
$endgroup$
For this problem, ignore A and C. You are just drawing $8$ bottles for B without replacement and need them all to be good. What is the chance the first bottle is good? Assuming it is, what is the chance the second is good? Keep going. The chances decrease because you are using up good bottles.
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
answered Jan 20 at 15:15
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
add a comment |
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