# Find $\mathbb{P}(A\cap B^c)$ where $A=\{X_1+X_2<1\}$ and $B=\{X_1+X_2+X_3<1\}$

Let $$(X_1,X_2,X_3)$$ three independent random variables with uniform distribution $$[0,1]$$.

• Let $$A=(X_1+X_2<1)$$. Find $$\mathbb{P}(A)$$.

$$\rightarrow \mathbb{P}(A)=\int_{0}^{1}[\int_{1-x_2}^{1}dx_1]dx_2=\frac{1}{2}$$.

• Let $$B=(X_1+X_2+X_3<1)$$. Find $$\mathbb{P}(B)$$.

$$\rightarrow \mathbb{P}(B)=\int_{0}^{1}dx_3[\int_{0}^{1-x_3}dx_2[\int_{0}^{1-x_3-x_2}dx_1]]=\frac{1}{6}$$

• Find $$\mathbb{P}(A\cap B^c)$$.

I think that first two points are correct, but I'm stuck at point 3). Anyone can help me? Thanks in advance.

Observe that for $$x_i \in (0,1)$$, if $$x_1+x_2+x_3 <1$$, then $$x_1+x_2<1$$ will automatically be satisfied. In other words, $$B \subseteq A$$. Thus $$A \cap B^c=A\setminus B$$. So the probability $$P(A \cap B^c)=P(A)-P(B)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$$.
If you draw the unit cube $$[0,1]^3$$, you will find that $$A$$ contains $$B$$ and that $$A\cap B^c$$ is just $$A - B$$. So the probability is $$1/2 - 1/6 = 1/3$$.