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

Question

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) $.

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

Answer 1

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 $P(A \cap B^c)=P(A)-P(B)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$.

Answer 2

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 $1/2 - 1/6 = 1/3$.