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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?

If I solve this question using arjun sir approach I'm getting $\frac{1}{16}$.

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

$=1 - \left( \dfrac{5}{18} + \dfrac{1}{36}\right)$
$= 1 - \dfrac{11}{36}$
$=\dfrac{25}{36}.$
by

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.

Because that has no significance (not favorable case). Say one friend arrives at 4:05 and other at 4:30, they don't meet.
@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.
@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)
Sir in the first case we multiply by 2 but in the second case, we do not by 2 why?

@Vaishali Trivedi

You can relate this to two coins being tossed. :)

In two coin tossings, the number of ways of getting a head and a tail = 2 -----{HT,TH}

But the number of ways of getting heads in both the coin tossings = 1 -----{HH}

Let us assume the origin to be equivalent to (4:00pm, 4:00pm)… and the marking on the x axis and y axis are in minutes…

$X$ is the random variable which shows the time instant at which friend $X$ arrives. And $Y$ is the random variable which shows the time interval at which friend $Y$ arrives.

Friend $Y$ shall not be able to meet $X$ if $Y$ arrives 10 mins after X arrives. So this is given by $\color{green}{Y>X+10}$ and the region is shown by $\color{green}{GREEN}$ in the graph above.

Similarly, Friend $X$ shall not be able to meet $Y$ if $X$ arrives 10 mins after Y arrives. So this is given by $\color{blue}{X>Y+10}$ and the region is shown by $\color{blue}{BLUE}$ in the graph above.

By virtue of how we have considered the picture above:

> they will wait for 10 minutes or the end of the hour whichever is earlier and leave.

Automatically gets covered…

So required probability = (Area in Blue+area in green)/Total area of the rectangle=25/36

@HitechGa nice way!!

could you please share some links of the resources from where i can learn where to apply this concept..?

we can also solve like this -

I thought that both approaches are inter-related and Arjun sir is also finding the required area using his calculation indirectly. But Now, I think both are different approaches. sorry bro I have wasted your time :(

brother I m not getting your solution ...can u plz explain in words?

Sir how you have drawn this area and how to think like this in exam

Simple trick

probability that one person meet on that day = 10/60 = 1/6
Probability(failing to meet) = 1-1/6=5/6
probability(failing to meet by both the persons) = 5/6*5/6= 25/36