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A man and a woman agree to meet at a certain location about $\text{12:30}$. If the man arrives at a time uniformly distributed between $\text{12:15}$ and $\text{12:45}$, and if the woman independently arrives at a time uniformly distributed between $\text{12:00}$ and $\text{13:00}$, find the probability that the first to arrive waits no longer than $5$ minutes. What is the probability that the man arrives first?

Surely I have problems to understand the limit of integration for the second part.

(a) Let $X$ be the time the man arrives and $Y$ the time the woman arrives (both in fractions of 1 hour) We want the probability $P(|X-Y|\le\frac{1}{12})$

The probability densities for each variable are $$f_x(t) = \begin{cases}2 &\frac{1}{4} \le t \le \frac{3}{4}\\ 0 & \text{otherwise}. \end{cases}$$ and

$$f_y(t) = \begin{cases}1 &0 \le t \le 1\\ 0 & \text{otherwise}. \end{cases}$$

The probability is given by $$P(|X-Y|\le\frac{1}{12})=P(Y-\frac{1}{12}\le X \le Y+\frac{1}{12})= \int_{s=\frac{1}{4}}^{s=\frac{3}{4}} \int_{y-\frac{1}{12}}^{y+\frac{1}{12}} f_x(t)\cdot f_y(t) dt~ds = \int_{s=\frac{1}{4}}^{s=\frac{3}{4}} \left(\int_{y-\frac{1}{12}}^{y+\frac{1}{12}} f_x(t)~dt\right)\,ds = \frac{1}{6}$$

(b) For the second part we have to calculate $P(X<Y)$ but I don't know what to do about the limits of integration. could someone help me?

I'm finding these kind of problems very difficult. Could you suggest me a strategy ?

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  • $\begingroup$ One would not want to mess up with $X$ and $x$, $Y$ and $y$. $\endgroup$ Commented Jan 5, 2017 at 14:37

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Are you simply asking for help with the second part?

We can think about it in the following way:

Can we compute the probability that $X$ is smaller than $t \in \mathbb{R}$? Can we compute the probability that $Y = t$? Then, we can solve the problem by conditioning.

$\mathbb{P}(X<Y) = \int_0^1 \mathbb{P}(X<Y|Y=t)f_Y(t)dt = \int_0^1 \mathbb{P}(X<t)dt$.

Now, split the last integral into the summation of integrals on sections on $[0,1/4), [1/4,3/4], (3/4,1]$. Can you solve it from here?

----------EDIT------------

So how do we determine $\mathbb{P}(X<t)$? We know that $X$ is uniformly distributed on $[1/4,3/4]$; therefore, it can never be less than $1/4$ and must always be less than $3/4$. Hence, for $t<1/4$, $\mathbb{P}(X<t) = 0$, and for $t>3/4$, $\mathbb{P}(X<t) = 1$. Now, for $t\in [1/4,3/4]$, we have that $\mathbb{P}(X<t)$ follows the cumulative distribution function (CDF) of a uniform distribution; hence,

$\mathbb{P}(X<t) = \frac{t-1/4}{3/4-1/4} = 2t-1/2$.

If you are unsure about the CDF of a uniform distribution, try to determine it by integration and by simple geometry.

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  • $\begingroup$ what is it your final risult? $\endgroup$ Commented Jan 5, 2017 at 15:28
  • $\begingroup$ the final answer should be 1/2 $\endgroup$ Commented Jan 5, 2017 at 15:35
  • $\begingroup$ Is this clear or do you need some more explicit detail? If it is sufficient, please upvote or mark as solved. Otherwise, feel free to ask for more detail. $\endgroup$ Commented Jan 5, 2017 at 16:04
  • $\begingroup$ From my calculation it is 1.are you sure that the first limits of integration are 0 and 1? $\endgroup$ Commented Jan 5, 2017 at 16:25
  • $\begingroup$ $\mathbb{P}(X<t) = \left\{\begin{matrix} 1, & t>3/4\\ 2t-1/2, &t\in [1/4,3/4] \\ 0, &t<1/4 \end{matrix}\right.$ Therefore we can split the integral into: $\int_0^1\mathbb{P}(X<t)dt = \int_{0}^{1/4} 0 dt + \int_{1/4}^{3/4}(2t-1/2)dt + \int_{3/4}^1 1 dt = 0 + 0.25 + 0.25$. $\endgroup$ Commented Jan 5, 2017 at 16:30

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