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This question was asked in one of sample paper. So I don't have an answer for this. It is as follows

Two friends A, B are planning to meet at the bus stop between 2 and 3 pm. The agreement between them is that after one arrives they will wait till other shows up, 15 minutes passed or it becomes 3 pm, whichever is earlier. Probability that they will both meet is ? (It had four options, A) 1/4 B) 3/16 C) 13/16 D) 3/4)

MY APPROACH

Let X, Y be the arrival times and we can assume that they are distributed as $U(0, 1)$ and are independent. So for them to meet we should have $|X - Y| < 1/4$.

region

As per the diagram the region of interest will be between the two lines. The lines are $x - y = \frac{1}{4}$ and $ y - x = \frac{1}{4}$. The area of triangle will be $ \frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}$. So the area of region will be $1 - 2 \times \frac{9}{32} = \frac{7}{16}$ which is not present in any of the option. So wanted to know if I am doing something wrong or was the options wrong ?

Amir
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redshellspy
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1 Answers1

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You are on the right track.

A more general version of the problem has been answered here using two methods:

Probability of no friends meeting at a coffee house

which shows that the probability of meeting of some friends is

$$\color{blue}{1- (1āˆ’(nāˆ’1)\delta)^n}.$$

In your case, $n=2$ and $\delta=\frac{15}{60}$, which gives

$$1- \left (1āˆ’(2āˆ’1)\frac{1}{4} \right)^2=\frac{7}{16}.$$

Other solution methods for $n=2$ can be found here.

Amir
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