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# Continuous Probability Question

Hi! Sharing a fun problem from SMO Senior 2013 on Continuous Probability :):

$$2$$ people go to the same swimming pool between $$2.00$$pm and $$5.00$$pm at random times and each swim for one hour. What is the chance that they will meet?

Note by Happy Melodies
3 years, 8 months ago

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Let $$2 \text{pm}$$ be the time $$\text{t} = 0$$,

Let one come at $$t_{1} = x \text{hrs.}$$ , and other at $$t_{2} = y \text{hrs.}$$

Then $$|x-y| \leq 1$$, $$\text{x,y} \leq 3$$

Let us make cases:

Case-1 $$x \leq 1$$, probability that $$t_{1} = x$$ is $$\frac{dx}{3}$$

Here, $$y \in (0 ,(x+1))$$ probability for which is $$\frac{x+1}{3}$$

Hence probability that $$x \leq 1$$ and $$y \in (0 ,(x+1))$$ is $$P_{1} = \displaystyle \int_{0}^{1} \frac{x+1}{9} \text{dx} = \frac{1}{6}$$

Case - 2 $$1 \leq x \leq 2$$ , probability that $$t_{1} = x$$ is $$\frac{dx}{3}$$

Here , $$y \in ((x-1),(x+1))$$ , probability for which is $$\frac{2}{3}$$

Hence probability that $$1 \leq x \leq 2$$ and $$y \in ((x-1) ,(x+1))$$ is $$P_{2} = \displaystyle \int_{1}^{2} \frac{2}{9} \text{dx} = \frac{2}{9}$$

Case- 3: Symmetrical to Case-1, $$P_{3} = \frac{1}{6}$$

Required probability = $$P_{1} + P_{2} + P_{3} = \frac{5}{9}$$ · 3 years, 8 months ago

Hi Jatin!

Can you please explain the following statement?

Case-1 $$x \leq 1$$, probability that $$t_{1} = x$$ is $$\frac{dx}{3}$$

Here, $$y \in (0 ,(x+1))$$ probability for which is $$\frac{x+1}{3}$$

How the probability is dx/3? I mean why do you need to introduce dx? Also, how do you find the range for y? I understand that this is a silly question but I have never seen anything like this before. Where can I find similar examples (or books which deal with this)?

Thanks! :) · 3 years, 8 months ago

Hi, Pranav,

Here, i am considering, One enters at time $$t_{1} \in (x - (x+dx))$$, then difference in probability for $$t_{1}= x$$ and $$t_{1} = x + dx$$ would be negligible.

Also, range for y is found by $$|x-y| \leq 1$$, which i had stated in solution. I don't know a book as such, but i have seen similar questions on brilliant such as:

Staying in the triangle · 3 years, 8 months ago

Problems like this often have a solution that interprets the probability as "area", mostly using coordinates. I will explain this.

Now, say we have the normal Cartesian plane, axes represent time, which units are 1 hour, with the origin representing noon for both x- and y-values(Draw it out!). If a point on the graph has coordinates (a,b), the x-value represents the time the first person went, the y-value for the second person. Let us look at the compliment of the two people meeting, that is, |a-b|>1. Suppose a>b, we have a-b>1, that is b<a-1. Now look at y<x-1, x<5, y>2, x>2, y<5, which are the conditions of the problem. The former 3 bounds a triangle that has an area of 2. We do the similar for b>a, that is b-a>1, which is b>a+1. Now look at y>x+1, x>2, y<5, x<5, y>2, which are the conditions of the problem. Again the former 3 bounds a triangle of area 2.

Now, probability of two people meeting is $$\frac{\text{Preferred outcomes}}{\text{Total outcomes}}=1-\frac{\text{Unpreferred outcomes}}{\text{Total outcomes}}=\frac{\text{Area of unpreferred outcomes}}{\text{Area of total outcomes}}$$. Total outcomes on the plane is the rectangle bounded by 2<x<5 and 2<y<5, which has area 9. Unpreferred outcomes is that they go to the swimming pool at durations more than 1 hour, which if the first person goes first, the area of that triangle is 2, and if the second person goes first, the area of that is 2, summing up to 4. Substituting the values, we get the probability $$\frac{5}{9}$$. · 3 years, 8 months ago