Probability problem on Unif(0,1) random variables

Question

Given three iid Unif(0,1) random variables a, b, and c, what is the probability that c is the maximum if a+b < 1?

Attempt: The probability distribution of a+b for a+b < 1 is a+b and for 1 < a+b < 2 it is 2-(a+b). Therefore, the events a+b < 1 and a+b > 1 both have probability 1/2. Also, $P(c>a \cap c>b) = 1/3$. This means that

$$1/3 = P(c>a \cap c>b | a+b > 1)P(a+b > 1) + P(c>a \cap c>b | a+b < 1)P(a+b < 1) $$

$$2/3 = P(c>a \cap c>b | a+b > 1) + P(c>a \cap c>b | a+b < 1) $$

I am not sure how to continue. Can someone explain how to proceed with the solution?

Answer 1

Representing the possible outcomes of sampling $a,b,c$ as three coordinates into a unit cube, the volume where $c>a,c>b,a+b<1$ is a triangular dipyramid with vertices $(0,0,0),(0,1,1),(1,0,1),(0,0,1),(1/2,1/2,1/2)$, which has base area $\frac{\sqrt3}2$ and combined height $\frac{\sqrt3}2$. Thus $P(c>a,c>b,a+b<1)=\frac13\cdot\frac{\sqrt3}2\cdot\frac{\sqrt3}2=\frac14$. More clearly, $P(a+b<1)=\frac12$ (a right triangular prism). Hence the final answer is $\frac{1/4}{1/2}=\frac12$.

Answer 2

The OP made a substantial change to is/her question. Here is another answer to the original posting.

Let $A=\{(a, b, c)\in(0,1)^3: a,b\leq c\}$ and $B=\{(a, b, c)\in(0,1)^3: a+b<1\}$. The qunatity of interest is $$P[A|B]=\frac{P[A\cap B]}{P[B]}$$ Now $P[B]=\int^1_0\int^{1-a}_0db\,da=\frac12$. Let is denote $x\wedge y=\min(x,y)$. Then

$$\begin{align} P[A\cap B]&=\int^1_0\int^c_0\int^{(1-a)\wedge c}_0 db \, da \, dc\\ &= \int^1_0\int^c_0(1-a)\wedge c\,da\,dc\\ &=\int^{1/2}_0\int^c_0(1-a)\wedge c\,da\,dc +\int^1_{1/2}\int^c_0(1-a)\wedge c\,da\,dc\\ &=\int^{1/2}_0c^2\,dc+\int^1_{1/2}\big(c-c^2 +\frac12(2c-1)\big)\,dc\\ &=\frac{1}{24}+(1-\frac14)-\frac13(1-\frac18)-\frac14\\ &=\frac{1}{24}+\frac{1}{2}-\frac{7}{24}=\frac14 \end{align} $$

Putting things together $$P[A|B]=\frac12$$