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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

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

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    One would not want to mess up with $X$ and $x$, $Y$ and $y$. – 2017-01-05

1 Answers 1

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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

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 $\mathbb{P}(X3/4$, $\mathbb{P}(X

$\mathbb{P}(X

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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    what is it your final risult? – 2017-01-05
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    the final answer should be 1/2 – 2017-01-05
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    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. – 2017-01-05
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    From my calculation it is 1.are you sure that the first limits of integration are 0 and 1? – 2017-01-05
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    $\mathbb{P}(X3/4\\ 2t-1/2, &t\in [1/4,3/4] \\ 0, &t<1/4 \end{matrix}\right.$ Therefore we can split the integral into: $\int_0^1\mathbb{P}(X – 2017-01-05
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    can you explain how you found $P(X – 2017-01-05
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    Sure, I'll edit my answer above with more detail. – 2017-01-05
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    please check what i've written here $$F_x(t) = P(x$$\int_{\frac{1}{4}}^{t}f_x(t)dt =\int_{\frac{1}{4}}^{t} 2 dt=2*(t-\frac{1}{4})=2*t-\frac{1}{2}$$ is it right? – 2017-01-05
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    @Anne That looks good. I'd be careful about using lower-case $x$ for a random variable (that is typically used to represent a realization of a random variable) - the preferred notation would be $X$. Similarly, the use of $*$ is typically reserved for the convolution operator, so I'd avoid that as well. The actual math looks correct though. – 2017-01-05