What is the probability $X + Y > 100$? $X,Y \sim U[40,80]$

Question

I'm struggling to write out the integral for this. Both $x$ and $y$ are i.i.d. distributed uniformly over $[40,80]$.

I think it should be of the form $\int_{40}^{80} \int_{100-x}^{80} \frac{1}{40\times40}dy~dx$ but this evaluates to $1$?

Thanks for your help!

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{40}^{80}{1 \over 40} \int_{40}^{80}{1 \over 40}\bracks{x + y > 100} \dd x\,\dd y} \\[5mm] = &\ {1 \over 1600}\int_{40}^{80}\int_{40}^{80} \bracks{x > 100 - y}\dd x\,\dd y \\[1cm] = &\ {1 \over 1600}\ \times \\[2mm] &\ \int_{40}^{80}\!\!\braces{\!\!% \bracks{100 - y < 40}\int_{40}^{80}\dd x + \bracks{40 < 100 - y < 80}\int_{100 - y}^{80}\dd x\!\!}\!\dd y \\[1cm] = &\ {1 \over 1600}\braces{% \int_{40}^{80}\bracks{y > 60}40\,\dd y + \int_{40}^{80} \bracks{y < 60}\pars{y - 20}\,\dd y } \\[5mm] = &\ {1 \over 40}\int_{60}^{80}\dd y + {1 \over 1600}\int_{40}^{60}\pars{y - 20}\dd y = \bbx{7 \over 8} = 0.875 \end{align}

Answer 2

The trick to these questions is to draw a picture. (In fact, I'd contend one should always attempt to draw a representative picture for math problems. The process of thinking about how to draw the picture is helpful to solve the problem.)

We're concerned with the region $x+y > 100$, where $x$ and $y$ are uniformly distributed from $40$ to $80$. So, let's plot $x+y=100$ and some bounds for $x=40$, $x=80$, $y=40$, and $y=80$:

Now, we return to your integral: \begin{align} P[x+y \ge 100\; |\; x,y \in A] &= \iint_A f_{X,Y}(x,y) \; dA \\ &= \iint_A f_X(x)\cdot f_Y(y) \; dA \quad \text{(because $X$ and $Y$ are iid)} \\ &= \iint_A \frac{1}{40}\cdot\frac{1}{40} \; dA \\ &= \frac{1}{1600} \iint_A \; dA \end{align}

Now, this raises the question --- how do we compute that integral? Fortunately, we don't have to! That integral represents the shaded area in the above graph. However, even that is kind-of nasty to find. Thus, we can use our complement rule and realize that the shaded area is just $1-(\text{probability represented by the missing corner})$.

That missing corner is just a triangle of base and height equal to $20$, so its area is $200$ square units. Let's denote that triangular region by $A^c$ (for $A$'s complement). Thus: \begin{align} P[x+y \ge 100\; |\; x,y \in A] &= \frac{1}{1600} \iint_A \; dA \\ &= 1 - \left(\frac{1}{1600} \iint_{A^c} \; dA^c \right) \\ &= 1 - \left(\frac{1}{1600} \cdot 200 \right) \\ &= 1 - \frac{1}{8} \\ &= \frac{7}{8} \end{align}

Answer 3

Comment. I used a simulation in R to make the picture suggested by @EthanBolker.

set.seed(1214);  m = 30000
x = runif(m, 40, 80);  y = runif(m, 40, 80)
s = x + y;  cond = (s > 100)
plot(x, y, pch=".")
points(x[cond], y[cond], pch=".", col="green2")
mean(cond)
[1] 0.8707

The distribution over the square is uniform. So it is enough to figure out what fraction of the square is colored green.

To make a nice picture $m = 30,000$ points is about right. But for a really good approximation to the right answer, not nearly enough. Even so, your answer ought to be somewhere near 0.87 (as elegantly shown by @apnorton).

Although it is not necessary to do so, it might be good practice to set up the limits for the double integral over the black region and subtract from $1.$ What is the (constant) value of the joint density function $f(x,y)$ over the square with vertices at $(40,40)$ and $(80,80)?$ Obviously, that's not completely trivial, because we accumulated one wrong answer on this page (now disappeared) while I was fussing with my simulation (and answering a couple of telemarketing calls).