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A and B decide to meet between 8 AM & 9 AM (60 minutes duration) at a specific location, and they have an agreement that no one will wait for other for more than 20 minutes. What is probability that, they'll meet?

I could not find out, how to approach this question? So, please guide me to appropriate topics.

David G. Stork
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g-217
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  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Matias Heikkilä Oct 03 '16 at 13:32

3 Answers3

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So, the conditions for meeting are: $$ \left\{ \begin{gathered} 0 \leqslant t_1 \leqslant 60 \hfill \\ 0 \leqslant t_2 \leqslant 60 \hfill \\ \left| {t_1 - t_2 } \right| \leqslant 20 \hfill \\ \end{gathered} \right. $$ which are geometrically represented as

meet
and it is just a matter of calculating the areas of triangles and square.

G Cab
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Draw a graph of the situation -- the $x$-axis is the time when $A$ arrives and the $y$-axis is the time when $B$ arrives, making the space of possibilities a square with side length $60$. The actual outcome that happens is represented by a uniformly chosen (or so we must assume) point in the square.

Now mark the area of the square where the two arrival times differ by no more than $20$. The probability of this happening is the ratio between the area you have marked and the area of the entire square.

hmakholm left over Monica
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    @gjha: no that's wrong: ($x-20$ gets values less than $0$ and $x+20$ more than $60$) – G Cab Oct 03 '16 at 15:52
  • Yeah .. it should be $\int_0^{40} (x + 20)dx + \int_{40}^{60} 60 dx - \int_{20}^{60} (x - 20)dx$ – g-217 Oct 04 '16 at 06:02
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Consider the positive coordinate axes. Let points on the positive x axis denote the times at which A arrives and the points on the y axis denote the times at which B arrives. Let the origin denote 8 a.m. 1 unit on each axis is equal to 1 hour. So all possibilities are included in the square S with opposite vertices (0,0) and (1,1) with 0 denoting 8 a.m and 1 denoting 9 a.m. For example, the event that A arrives at 8:30 a.m and B arrives at 8:45 a.m is denoted by the point (1/2, 3/4).

Let x be the time at which A arrives and y be the time at which B arrives. The constraints are x-y < 1/3 and y-x < 1/3 with 1/3 representing 20 minutes (1/3 hour). Draw these lines on the square S. The area enclosed between these lines and the square divided by the area of the square is the required probability. 5/9 is the answer.

You can also work this out using calculus. Assume arrival times in a unit time interval representing the time period from 8 a.m to 9 p.m follow a uniform distribution. Suppose A arrives at time x then B must arrive within 20 minutes (1/3 hour) before or after A for them to meet. The chance of this happening depends on x. There are three cases-

x < 1/3, 1/3 <= x < 2/3, 2/3 <= x < 1

Case 1: B must arrive in the first x + 1/3 of [0,1] interval. Probability of this is (x + 1/3)/1 (uniform distribution). So probability that A and B meet given A arrives at x is (x + 1/3) dx.

Case 2: B must arrive within 1/3 hour before or after x. Probability that A and B meet given A arrives at x is 2/3 dx.

Case 3: B must arrive within the last 1/3 + 1 - x hours. Probability that A and B meet given A arrives at x is (1/3 + 1 -x) dx.

Integrate each of the expressions between the appropriate limits (Case 1: 0 to 1/3), (Case 2: 1/3 to 2/3) and (Case 3: 2/3 to 1) and add the integrals to get the answer.

user2371765
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