Probability that friends meet.
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Two friends decide to meet between 1PM and 2PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. What is the probability that they meet ?
How to solve this problem using Random Variables mathematically ?
probability probability-distributions
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
$endgroup$
add a comment |
$begingroup$
Two friends decide to meet between 1PM and 2PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. What is the probability that they meet ?
How to solve this problem using Random Variables mathematically ?
probability probability-distributions
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
$endgroup$
$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
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– Cato
Jan 26 '17 at 17:49
$begingroup$
related math.stackexchange.com/questions/103015/…
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– Henry
May 7 '17 at 18:55
add a comment |
$begingroup$
Two friends decide to meet between 1PM and 2PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. What is the probability that they meet ?
How to solve this problem using Random Variables mathematically ?
probability probability-distributions
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
$endgroup$
Two friends decide to meet between 1PM and 2PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. What is the probability that they meet ?
How to solve this problem using Random Variables mathematically ?
probability probability-distributions
probability probability-distributions
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
asked Jan 26 '17 at 17:42
S PaulS Paul
10811
10811
$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
$endgroup$
– Cato
Jan 26 '17 at 17:49
$begingroup$
related math.stackexchange.com/questions/103015/…
$endgroup$
– Henry
May 7 '17 at 18:55
add a comment |
$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
$endgroup$
– Cato
Jan 26 '17 at 17:49
$begingroup$
related math.stackexchange.com/questions/103015/…
$endgroup$
– Henry
May 7 '17 at 18:55
$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
$endgroup$
– Cato
Jan 26 '17 at 17:49
$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
$endgroup$
– Cato
Jan 26 '17 at 17:49
$begingroup$
related math.stackexchange.com/questions/103015/…
$endgroup$
– Henry
May 7 '17 at 18:55
$begingroup$
related math.stackexchange.com/questions/103015/…
$endgroup$
– Henry
May 7 '17 at 18:55
add a comment |
1 Answer
1
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One man must arrive before the other. The probability that man 1 arrives during the first $frac34$ hour is $frac34$. He'll then wait $frac14$ hour.
The probability that he arrives during the last $frac14$ hour is $frac14$, and then (on average he'll wait) $frac18$ hour.
So altogether the man 1 will wait $frac34 × frac14 + frac14 × frac18 = frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $frac7{32} + frac7{32} = frac7{16}$
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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active
oldest
votes
$begingroup$
One man must arrive before the other. The probability that man 1 arrives during the first $frac34$ hour is $frac34$. He'll then wait $frac14$ hour.
The probability that he arrives during the last $frac14$ hour is $frac14$, and then (on average he'll wait) $frac18$ hour.
So altogether the man 1 will wait $frac34 × frac14 + frac14 × frac18 = frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $frac7{32} + frac7{32} = frac7{16}$
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
$endgroup$
add a comment |
$begingroup$
One man must arrive before the other. The probability that man 1 arrives during the first $frac34$ hour is $frac34$. He'll then wait $frac14$ hour.
The probability that he arrives during the last $frac14$ hour is $frac14$, and then (on average he'll wait) $frac18$ hour.
So altogether the man 1 will wait $frac34 × frac14 + frac14 × frac18 = frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $frac7{32} + frac7{32} = frac7{16}$
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
$endgroup$
add a comment |
$begingroup$
One man must arrive before the other. The probability that man 1 arrives during the first $frac34$ hour is $frac34$. He'll then wait $frac14$ hour.
The probability that he arrives during the last $frac14$ hour is $frac14$, and then (on average he'll wait) $frac18$ hour.
So altogether the man 1 will wait $frac34 × frac14 + frac14 × frac18 = frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $frac7{32} + frac7{32} = frac7{16}$
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
$endgroup$
One man must arrive before the other. The probability that man 1 arrives during the first $frac34$ hour is $frac34$. He'll then wait $frac14$ hour.
The probability that he arrives during the last $frac14$ hour is $frac14$, and then (on average he'll wait) $frac18$ hour.
So altogether the man 1 will wait $frac34 × frac14 + frac14 × frac18 = frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $frac7{32} + frac7{32} = frac7{16}$
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
answered Jan 26 '17 at 18:10
Kanwaljit SinghKanwaljit Singh
8,5351517
8,5351517
add a comment |
add a comment |
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$begingroup$
you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
$endgroup$
– Cato
Jan 26 '17 at 17:49
$begingroup$
related math.stackexchange.com/questions/103015/…
$endgroup$
– Henry
May 7 '17 at 18:55