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+12 votes
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Two friends agree to meet at a park with the following conditions. Each will reach the park between 4:00 pm and 5:00 pm and will see if the other has already arrived. If not, they will wait for 10 minutes or the end of the hour whichever is earlier and leave. What is the probability that the two will not meet?
asked in Probability by Veteran (59.5k points) | 891 views
+2
http://math.stackexchange.com/a/1279931/309722
Here answer is given as $\frac{7}{16}$
0
In that Link Que, Offset is 15 Mins. Hence answer is diferent. Here offset is 10 mins.
0
applying above method answer came as 25/36.

2 Answers

+13 votes
Best answer
We are given that both will be reaching the park between $4:00$ and $5:00$.

Probability that one friend arrives between $4:00$ and $4:50 = \dfrac{5}{6}$
Probability that one friend arrives between $4:00$ and $4:50$ and meets the other arriving in the next $10$ minutes $=\dfrac{5}{6}\times \dfrac{1}{6}\times {2} =\dfrac{10}{36} =\dfrac{5}{18}.$
(For any time of arrival between $4:00$ and $4:50,$ we have a $10$ minute interval possible for the second friend to arrive, and $2$ cases as for choosing which friend arrives first)

Probability that both friend arrives between $4:50$ and $5:00 =\dfrac{1}{6}\times \dfrac{1}{6} =\dfrac{1}{36}.$

This covers all possibility of a meet. So, required probability of non-meet

$=1 - \left( \dfrac{5}{18} + \dfrac{1}{36}\right)$
$= 1 - \dfrac{11}{36}$
$=\dfrac{25}{36}.$
answered by Veteran (355k points)
edited by
0
why did not u consider the case when both of them arrive between 4.00 pm and 4.50 pm?

Instead u considered the case where both of them arrived between 4.50 pm and 5 pm.

Please explain.
0
Because that has no significance (not favorable case). Say one friend arrives at 4:05 and other at 4:30, they don't meet.
0
@Arjun sir, Suppose one come @10:30 another @10:35 then they will meet right?? bcoz the one who came earlier will wait 10 mins.

but this case is not considered here.
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@Rajesh Pradhan This case is included when he says the following:
The probability that one friend arrives between 4:00 and 04:50 and meets the other arriving in the next 1010 minutes =56×16×2=1036=518.=56×16×2=1036=518.
(For any time of arrival between 4:00 and 4:50, we have a 10 minute interval possible for the second friend to arrive, and 2 cases as for choosing which friend arrives first)
+1 vote

we can also solve like this -

answered by Loyal (5.3k points)


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