Mixing
uniformly distributed random variables
$begingroup$
Ram and Shyam wanted to meet at a park about 12.30 P.M.. If Ram arrives at a time uniformly distributed between 12.15 P.M. to 12.45 P.M. and if Shyam independently arrives at a time uniformly distributed between 12.00 P.M. and 1.00 P.M., then find the probability that the first to arrive waits no longer than 5 minutes
probability-distributions uniform-distribution
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
$endgroup$
add a comment |
$begingroup$
Ram and Shyam wanted to meet at a park about 12.30 P.M.. If Ram arrives at a time uniformly distributed between 12.15 P.M. to 12.45 P.M. and if Shyam independently arrives at a time uniformly distributed between 12.00 P.M. and 1.00 P.M., then find the probability that the first to arrive waits no longer than 5 minutes
probability-distributions uniform-distribution
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
$endgroup$
$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33
add a comment |
$begingroup$
Ram and Shyam wanted to meet at a park about 12.30 P.M.. If Ram arrives at a time uniformly distributed between 12.15 P.M. to 12.45 P.M. and if Shyam independently arrives at a time uniformly distributed between 12.00 P.M. and 1.00 P.M., then find the probability that the first to arrive waits no longer than 5 minutes
probability-distributions uniform-distribution
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
$endgroup$
Ram and Shyam wanted to meet at a park about 12.30 P.M.. If Ram arrives at a time uniformly distributed between 12.15 P.M. to 12.45 P.M. and if Shyam independently arrives at a time uniformly distributed between 12.00 P.M. and 1.00 P.M., then find the probability that the first to arrive waits no longer than 5 minutes
probability-distributions uniform-distribution
probability-distributions uniform-distribution
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
edited Apr 17 '15 at 9:18
Monica Jain
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
asked Apr 8 '15 at 13:59
Monica JainMonica Jain
62
62
$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33
add a comment |
$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33
$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute.
Ram has to come from 12.15 to 12.45
Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes
As per the rule of uniform distribution
P = 1/(b-a). f(X) dx
So sample space would be time from 12.10 P.M to 12.50 PM
Therefore, a = 10, b = 5
P = 1/(50-10). f(X)
P =1/(40) . (0-5)
P = 5/40 = 1/8
Probability (P) = 1/8
Ans : So probability of first to arrive wait no longer than 5 minute would be 1/8.
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
$endgroup$
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute.
Ram has to come from 12.15 to 12.45
Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes
As per the rule of uniform distribution
P = 1/(b-a). f(X) dx
So sample space would be time from 12.10 P.M to 12.50 PM
Therefore, a = 10, b = 5
P = 1/(50-10). f(X)
P =1/(40) . (0-5)
P = 5/40 = 1/8
Probability (P) = 1/8
Ans : So probability of first to arrive wait no longer than 5 minute would be 1/8.
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
$endgroup$
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
add a comment |
$begingroup$
Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute.
Ram has to come from 12.15 to 12.45
Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes
As per the rule of uniform distribution
P = 1/(b-a). f(X) dx
So sample space would be time from 12.10 P.M to 12.50 PM
Therefore, a = 10, b = 5
P = 1/(50-10). f(X)
P =1/(40) . (0-5)
P = 5/40 = 1/8
Probability (P) = 1/8
Ans : So probability of first to arrive wait no longer than 5 minute would be 1/8.
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
$endgroup$
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
add a comment |
$begingroup$
Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute.
Ram has to come from 12.15 to 12.45
Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes
As per the rule of uniform distribution
P = 1/(b-a). f(X) dx
So sample space would be time from 12.10 P.M to 12.50 PM
Therefore, a = 10, b = 5
P = 1/(50-10). f(X)
P =1/(40) . (0-5)
P = 5/40 = 1/8
Probability (P) = 1/8
Ans : So probability of first to arrive wait no longer than 5 minute would be 1/8.
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
$endgroup$
Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute.
Ram has to come from 12.15 to 12.45
Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes
As per the rule of uniform distribution
P = 1/(b-a). f(X) dx
So sample space would be time from 12.10 P.M to 12.50 PM
Therefore, a = 10, b = 5
P = 1/(50-10). f(X)
P =1/(40) . (0-5)
P = 5/40 = 1/8
Probability (P) = 1/8
Ans : So probability of first to arrive wait no longer than 5 minute would be 1/8.
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
answered Apr 10 '15 at 5:01
Monica JainMonica Jain
62
62
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
add a comment |
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
$begingroup$
Hi all, I would still like few comments on this solution. If it is accurate. Thanks
$endgroup$
– Monica Jain
Apr 10 '15 at 5:02
add a comment |
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$begingroup$
similar question math.stackexchange.com/questions/1218478/…
$endgroup$
– Monica Jain
Apr 8 '15 at 14:00
$begingroup$
An answer to a very similar problem can be found here. Think how you can adapt this to your problem. Here the problem is solved in a geometric (visual) fashion, which is easy to get.
$endgroup$
– Pedro
Apr 8 '15 at 14:47
$begingroup$
A friend who randomly arrives between 12 and 1 ... is not a friend you want. Hint: find a friend who can keep an appointment.
$endgroup$
– wolfies
Apr 8 '15 at 15:13
$begingroup$
hi Pedro, the link you shared is very good but that is for continuous distribution and not uniform distribution.
$endgroup$
– Monica Jain
Apr 9 '15 at 11:33