4
$\begingroup$

$A$ and $B$ decide to meet at a cafe between $5$ p.m. and $6$ p.m. They agree that the person who arrives first at the cafe would wait for exactly $15$ minutes for the other. If each of them arrives at a random time between $5$ p.m. and $6$ p.m., what is the probability that the meeting takes place?

I figured that if one of them arrive at the first minute then the probability of the two meeting each other would be $15/60$, because the second person could arrive from the $1^{st}$ minute till the $15^{th}$ minute and meet with him. Similarly if the first person arrives at the second minute the probability would be $16/60$. This will go on till the $14^{th}$ minute and the probability would be $29/60$. The probability will remain $29/60$ till the $45^{th}$ minute, after which it will gradually decrease in the order $28/60, 27/60,... , 15/60.$

I am not sure if my approach is correct. Also I am stuck after a point with my approach. Please explain elaborately how to solve such questions.

$\endgroup$
1
$\begingroup$

Let $X$ and $Y$ be the times in units of hours that $X$ and $Y$ arrive. I assume here that they are uniformly distributed on $[0,1]$ and independent. Then the meeting happens provided $|X-Y| \leq 1/4$. So the probability of the meeting is

$$\frac{\int_{|x-y| \leq 1/4,0 \leq x \leq 1,0 \leq y \leq 1} dx dy}{\int_{0 \leq x \leq 1,0 \leq y \leq 1} dx dy}.$$

That is, it is the area of the region in the plane where they meet divided by the area of the square (which is just $1$). This region is the square except for the two triangles which lie above $y=x+1/4$ and below $y=x-1/4$. These have height and width $3/4$, so their areas are each $9/32$, which add up to $9/16$. So the area of the region is $7/16$, which is also the probability of the meeting.

A similar argument can be done when you assume that $X$ and $Y$ have a discrete distribution instead (as you seem to be doing in the original question).

$\endgroup$
  • $\begingroup$ The area of the plane where they meet is not a parallelogram, but an irregular hexagon. "Base times height" doesn't work for that. $\endgroup$ – hmakholm left over Monica May 10 '15 at 14:10
  • $\begingroup$ I see, I missed the sides of the polygon on the vertical sides of the square. Let me try to work on it. $\endgroup$ – Ian May 10 '15 at 14:13
  • 1
    $\begingroup$ Looking at the area is the easiest way of doing it, but instead of finding the area of the hexagon directly, it is easiest to find the areas of the two triangles outside the hexagon and then subtract them from the whole square. $\endgroup$ – hmakholm left over Monica May 10 '15 at 14:14
  • $\begingroup$ @Ian - The answer that is given is 7/16 . $\endgroup$ – riz May 10 '15 at 14:15
  • $\begingroup$ @HenningMakholm Nice trick there. $\endgroup$ – Ian May 10 '15 at 14:17
1
$\begingroup$

A graphical solution to this problem:

Let A and B be Alice and Bob's arrival times, both variables taking on values between 0 and 1.

Since Alice and Bob must arrive within 1/4 hours of each other, the following equation must be satisfied:

$$|A - B| \leq 1/4$$

We can break apart this absolute value into the following two cases:

$$ A \geq B + 1/4, and, A \leq B - 1/4$$

Graphing these inequalities on a plot where A and B are between 0 and 1, creates three regions. The region where the inequality is satisfied is the diagonal strip down the middle.

Since probability is proportional to area, and the total area of the plot is 1, the area of the diagonal strip is the probability that the meeting will take place. Calculating that probability:

$$ P = 1 - (3/4)^2 = 7/16$$

$\endgroup$
  • $\begingroup$ $|A-B|\le\frac14$ does not "break into two cases". There is exactly one case: $$A\ge B-\frac14\wedge A\le B+\frac14 $$ $\endgroup$ – user228113 Nov 16 '15 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.