Probability that Carl has to wait at least 10 minutes for one of the others and for both of the others to...











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This question is for the other subquestions for the same problem here. For those not willing to click the link, I will post the exercise problem here as well.




Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m.



For the rest of the problem, assume Carl arrives first at 1:10 p.m. and condition on this fact.



(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?



(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?




Apparently, my approach to the solution has given me the answers in a swapped manner, which I can't understand why. Allow me to elaborate.






My approach



Let's say that the events for Alice and Bob arriving are each $A$ and $B$ i.i.d. on Unif($10, 30$).






(b) The probability that Carl will wait more than 10 minutes for at least one of the others to arrive is:



(Probability that Carl waits less than 10 minutes for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both)



which gives me the result of $(frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) = frac{3}{4}$.






(c) Probability that Carl waits more than 10 minutes for both is the probability that $A$ and $B$ both fall in the interval ($20$, $30$) in the total interval ($10$, $30$). Therefore we get the probability $frac{1}{2} * frac{1}{2} = frac{1}{4}$.






However, apparently the correct answer is $frac{1}{4}$ for (b) and $frac{3}{4}$ for (c), and I'm not entirely sure why.



The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $1 - frac{1}{4}$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight.






Thank you.






probability probability-theory uniform-distribution







share|cite|improve this question




















  • 1




    The explanations in the book seem reversed to me. I get the same answers you do.
    – saulspatz
    Nov 25 at 4:04










  • That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
    – Sean
    Nov 25 at 4:26










  • I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
    – Sauhard Sharma
    Nov 25 at 4:28






  • 1




    Thanks. Then your solution looks good.
    – Sauhard Sharma
    Nov 25 at 4:36






  • 1




    I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
    – saulspatz
    Nov 25 at 6:57















up vote
3
down vote

favorite












This question is for the other subquestions for the same problem here. For those not willing to click the link, I will post the exercise problem here as well.




Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m.



For the rest of the problem, assume Carl arrives first at 1:10 p.m. and condition on this fact.



(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?



(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?




Apparently, my approach to the solution has given me the answers in a swapped manner, which I can't understand why. Allow me to elaborate.






My approach



Let's say that the events for Alice and Bob arriving are each $A$ and $B$ i.i.d. on Unif($10, 30$).






(b) The probability that Carl will wait more than 10 minutes for at least one of the others to arrive is:



(Probability that Carl waits less than 10 minutes for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both)



which gives me the result of $(frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) = frac{3}{4}$.






(c) Probability that Carl waits more than 10 minutes for both is the probability that $A$ and $B$ both fall in the interval ($20$, $30$) in the total interval ($10$, $30$). Therefore we get the probability $frac{1}{2} * frac{1}{2} = frac{1}{4}$.






However, apparently the correct answer is $frac{1}{4}$ for (b) and $frac{3}{4}$ for (c), and I'm not entirely sure why.



The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $1 - frac{1}{4}$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight.






Thank you.






probability probability-theory uniform-distribution







share|cite|improve this question




















  • 1




    The explanations in the book seem reversed to me. I get the same answers you do.
    – saulspatz
    Nov 25 at 4:04










  • That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
    – Sean
    Nov 25 at 4:26










  • I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
    – Sauhard Sharma
    Nov 25 at 4:28






  • 1




    Thanks. Then your solution looks good.
    – Sauhard Sharma
    Nov 25 at 4:36






  • 1




    I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
    – saulspatz
    Nov 25 at 6:57













up vote
3
down vote

favorite









up vote
3
down vote

favorite











This question is for the other subquestions for the same problem here. For those not willing to click the link, I will post the exercise problem here as well.




Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m.



For the rest of the problem, assume Carl arrives first at 1:10 p.m. and condition on this fact.



(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?



(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?




Apparently, my approach to the solution has given me the answers in a swapped manner, which I can't understand why. Allow me to elaborate.






My approach



Let's say that the events for Alice and Bob arriving are each $A$ and $B$ i.i.d. on Unif($10, 30$).






(b) The probability that Carl will wait more than 10 minutes for at least one of the others to arrive is:



(Probability that Carl waits less than 10 minutes for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both)



which gives me the result of $(frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) = frac{3}{4}$.






(c) Probability that Carl waits more than 10 minutes for both is the probability that $A$ and $B$ both fall in the interval ($20$, $30$) in the total interval ($10$, $30$). Therefore we get the probability $frac{1}{2} * frac{1}{2} = frac{1}{4}$.






However, apparently the correct answer is $frac{1}{4}$ for (b) and $frac{3}{4}$ for (c), and I'm not entirely sure why.



The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $1 - frac{1}{4}$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight.






Thank you.






probability probability-theory uniform-distribution







share|cite|improve this question















This question is for the other subquestions for the same problem here. For those not willing to click the link, I will post the exercise problem here as well.




Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m.



For the rest of the problem, assume Carl arrives first at 1:10 p.m. and condition on this fact.



(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?



(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?




Apparently, my approach to the solution has given me the answers in a swapped manner, which I can't understand why. Allow me to elaborate.






My approach



Let's say that the events for Alice and Bob arriving are each $A$ and $B$ i.i.d. on Unif($10, 30$).






(b) The probability that Carl will wait more than 10 minutes for at least one of the others to arrive is:



(Probability that Carl waits less than 10 minutes for Alice and more than 10 for Bob) + (Probability that Carl waits less than 10 minutes for Bob and more than 10 for Alice) + (Probability that Carl waits more than 10 minutes for both)



which gives me the result of $(frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) + (frac{1}{2} * frac{1}{2}) = frac{3}{4}$.






(c) Probability that Carl waits more than 10 minutes for both is the probability that $A$ and $B$ both fall in the interval ($20$, $30$) in the total interval ($10$, $30$). Therefore we get the probability $frac{1}{2} * frac{1}{2} = frac{1}{4}$.






However, apparently the correct answer is $frac{1}{4}$ for (b) and $frac{3}{4}$ for (c), and I'm not entirely sure why.



The rationale the textbook gives is that for (b) we need to compute the time that both of them arrive after 1:20 p.m., and for (c) we simple take the complement that both of them arrive between 1:10 and 1:20 p.m. $1 - frac{1}{4}$. I'm not entirely sure how these solutions make sense, though, and was hoping someone would be kind enough to provide me with some insight.






Thank you.






probability probability-theory uniform-distribution




probability probability-theory uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 4:24

























asked Nov 25 at 3:56









Sean

23510




23510








  • 1




    The explanations in the book seem reversed to me. I get the same answers you do.
    – saulspatz
    Nov 25 at 4:04










  • That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
    – Sean
    Nov 25 at 4:26










  • I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
    – Sauhard Sharma
    Nov 25 at 4:28






  • 1




    Thanks. Then your solution looks good.
    – Sauhard Sharma
    Nov 25 at 4:36






  • 1




    I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
    – saulspatz
    Nov 25 at 6:57














  • 1




    The explanations in the book seem reversed to me. I get the same answers you do.
    – saulspatz
    Nov 25 at 4:04










  • That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
    – Sean
    Nov 25 at 4:26










  • I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
    – Sauhard Sharma
    Nov 25 at 4:28






  • 1




    Thanks. Then your solution looks good.
    – Sauhard Sharma
    Nov 25 at 4:36






  • 1




    I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
    – saulspatz
    Nov 25 at 6:57








1




1




The explanations in the book seem reversed to me. I get the same answers you do.
– saulspatz
Nov 25 at 4:04




The explanations in the book seem reversed to me. I get the same answers you do.
– saulspatz
Nov 25 at 4:04












That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
– Sean
Nov 25 at 4:26




That's strange... I should bring this issue up with my instructor again, as the first time it was brushed off. May I ask, was the rationale that led you to your answer the same as mine?
– Sean
Nov 25 at 4:26












I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
– Sauhard Sharma
Nov 25 at 4:28




I'm getting the answers as $frac{1}{3}$ for $b)$ and $frac{1}{9}$ for $c)$. Also how can you change the probability distribution of Alice and Bob from $Uniform(0,30)$ to $Uniform(10,30)$
– Sauhard Sharma
Nov 25 at 4:28




1




1




Thanks. Then your solution looks good.
– Sauhard Sharma
Nov 25 at 4:36




Thanks. Then your solution looks good.
– Sauhard Sharma
Nov 25 at 4:36




1




1




I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
– saulspatz
Nov 25 at 6:57




I did it exactly as the book describes, but for the other problems. I think the answers are simply reversed.
– saulspatz
Nov 25 at 6:57















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