Probability that at least two of three uniform random variables~[0 1.5] add up to >2 There's a problem I've been stuck on for a while regarding the sum of two uniformly distributed, independent random variables. The problem goes like this:
You find some old batteries in a drawer. They produce the voltage Xi. Assume that Xi i.i.d. and uniformly distributed in [0, 1.5].
You pick three batteries. What is the probability that at least two of them have voltages that add up to more than 2V?
I've been able to find the probability that any set of two of them add up to more than 2, but I can't figure out how to set up the rest. Thanks in advance!
 A: Let's rescale to $[0,1]$ and a sum of $\frac43$ to make it a bit easier.
Fleshing out Brian's suggestion:
At least one pair sums to at least $\frac43$ if the maximum pair sums to at least $\frac43$.
The maximum voltage has cumulative distribution function $x^3$ and thus probability density function $3x^2$. Conditional on $x$, the second-highest voltage has cumulative distribution function $\frac{y^2}{x^2}$ and thus probability density function $\frac{2y}{x^2}$. For the sum to be at least $\frac43$, the maximum has to be at least $\frac23$. The desired probability is
$$
\int_\frac23^1\mathrm dx\,3x^2\int_{\frac43-x}^x\mathrm dy\frac{2y}{x^2}=6\int_\frac23^1\mathrm dx\int_{\frac43-x}^x\mathrm dy\,y\;,
$$
which by the way can also be interpreted as
$$
3!\int_\frac23^1\mathrm dx\int_{\frac43-x}^x\mathrm dy\int_0^y\mathrm dz\;,
$$
with a fixed order for the three variables and a factor $3!$ for the possible orders, another way to set up the integration. In any case, the result is
$$
6\int_\frac23^1\mathrm dx\int_{\frac43-x}^x\mathrm dy\,y=3\int_\frac23^1\mathrm dx\left(x^2-\left(\frac43-x\right)^2\right)=\frac49\;.
$$
