Confuse in Probability in Discrete Mathematics about train late problem
$begingroup$
A student is being late to catch the morning train in 3/10 of trials. If he runs of on time he will always catch the train. If he runs of late, he gets to the station with 5 min delay after regular train departure and so he catches the train if the train is at least 5 min delayed.
The train is at least 5 min delayed with probability 2/5. The student is at school on time in case that the train gets to his school with delay at most 30 min.
Probability that the delay of the train in the city he lived in is less than 30 minutes provided student at his station got on train on time (with delay less than 5 min) is 99/100.
Probability that the delay of the train in that city is less than 30 minutes provided student got on train with delay more than 5 min is 97/100.
What is the probability, that student is at school on time on a day when he runs of late?
I have read this problem for many time but i still can't figure out how to solve it, there are many infomation and it make me confuse, i don't know where to start and which infomation that i should be focus on.
Please show steps in how you solve it.
probability discrete-mathematics
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
$endgroup$
add a comment |
$begingroup$
A student is being late to catch the morning train in 3/10 of trials. If he runs of on time he will always catch the train. If he runs of late, he gets to the station with 5 min delay after regular train departure and so he catches the train if the train is at least 5 min delayed.
The train is at least 5 min delayed with probability 2/5. The student is at school on time in case that the train gets to his school with delay at most 30 min.
Probability that the delay of the train in the city he lived in is less than 30 minutes provided student at his station got on train on time (with delay less than 5 min) is 99/100.
Probability that the delay of the train in that city is less than 30 minutes provided student got on train with delay more than 5 min is 97/100.
What is the probability, that student is at school on time on a day when he runs of late?
I have read this problem for many time but i still can't figure out how to solve it, there are many infomation and it make me confuse, i don't know where to start and which infomation that i should be focus on.
Please show steps in how you solve it.
probability discrete-mathematics
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
$endgroup$
$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29
add a comment |
$begingroup$
A student is being late to catch the morning train in 3/10 of trials. If he runs of on time he will always catch the train. If he runs of late, he gets to the station with 5 min delay after regular train departure and so he catches the train if the train is at least 5 min delayed.
The train is at least 5 min delayed with probability 2/5. The student is at school on time in case that the train gets to his school with delay at most 30 min.
Probability that the delay of the train in the city he lived in is less than 30 minutes provided student at his station got on train on time (with delay less than 5 min) is 99/100.
Probability that the delay of the train in that city is less than 30 minutes provided student got on train with delay more than 5 min is 97/100.
What is the probability, that student is at school on time on a day when he runs of late?
I have read this problem for many time but i still can't figure out how to solve it, there are many infomation and it make me confuse, i don't know where to start and which infomation that i should be focus on.
Please show steps in how you solve it.
probability discrete-mathematics
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
$endgroup$
A student is being late to catch the morning train in 3/10 of trials. If he runs of on time he will always catch the train. If he runs of late, he gets to the station with 5 min delay after regular train departure and so he catches the train if the train is at least 5 min delayed.
The train is at least 5 min delayed with probability 2/5. The student is at school on time in case that the train gets to his school with delay at most 30 min.
Probability that the delay of the train in the city he lived in is less than 30 minutes provided student at his station got on train on time (with delay less than 5 min) is 99/100.
Probability that the delay of the train in that city is less than 30 minutes provided student got on train with delay more than 5 min is 97/100.
What is the probability, that student is at school on time on a day when he runs of late?
I have read this problem for many time but i still can't figure out how to solve it, there are many infomation and it make me confuse, i don't know where to start and which infomation that i should be focus on.
Please show steps in how you solve it.
probability discrete-mathematics
probability discrete-mathematics
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
edited Dec 11 '18 at 18:30
Alex Ravsky
40.4k32282
40.4k32282
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
asked Dec 9 '18 at 20:57
Hieu NgHieu Ng
1
1
$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29
add a comment |
$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29
$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What is the probability, that student is at school on time on a day when he runs of late?
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
add a comment |
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$begingroup$
What is the probability, that student is at school on time on a day when he runs of late?
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
add a comment |
$begingroup$
What is the probability, that student is at school on time on a day when he runs of late?
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
add a comment |
$begingroup$
What is the probability, that student is at school on time on a day when he runs of late?
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
What is the probability, that student is at school on time on a day when he runs of late?
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 11 '18 at 18:26
Alex RavskyAlex Ravsky
40.4k32282
40.4k32282
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
add a comment |
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
1
1
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
$begingroup$
Thank you so much, i think i should think more simple about this problem.
$endgroup$
– Hieu Ng
Dec 17 '18 at 7:21
add a comment |
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$begingroup$
Hello and welcome to Mathematics SE. Can you show us what you have tried so far? It might help you to make a flowchart with scenarios. So start with is the student late or not. Then if he is delayed, is the train delayed or not? And so on. This will help you get an overview and likely allows you to solve the problem without giving away the solution.
$endgroup$
– Jan
Dec 9 '18 at 20:59
$begingroup$
So the problem ask me to count the probability that the student can be at school on time on the day he runs of late, so: - If the student runs of late he will have 2/5 change to get on train. - Students can get to school on time if the train can not be delay more than 30 min: + The student will have 1/100 chance to get on train and get to school on time if the train is delay less than 5 min. + The student will have 3/100 chance to get on train and get to school on time if the train is delay more than 5 min. After all, i really confuse and don't know how to connect these data together
$endgroup$
– Hieu Ng
Dec 9 '18 at 22:13
$begingroup$
It seems that the conditions after '+'s are in mess, they should be about train delay at least 30 minutes.
$endgroup$
– Alex Ravsky
Dec 11 '18 at 18:29