1. ## Two proofs (probability)

Hello

Help me prove these two classical rules

1. A and B are two events. Show that the probability that exactly one of these events will occur is $P(A) + P(B) - 2P(A \cap B)$

2. We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Prove that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

Thank you.

2. Originally Posted by Tau
1. A and B are two events. Show that the probability that exactly one of these events will occur is $P(A) + P(B) - 2P(A \cap B)$

2. We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Prove that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$
Exactly one is: $P(A\cap B^c)+P(B\cap A^c)$.
Recall that $P(A\cap B^c)=P(A)-P(A\cap B).$

In #2 let $D=A\cup B$. The expand $P(D\cup C)$.
The go back and make subsitutions.

3. Hello, Tau!

2. We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Prove that: . $P(A \cup B \cup C) \; = \;\begin{Bmatrix} P(A) + P(B) + P(C) \\
- P(A \cap B) - P(A \cap C) - P(B \cap C) \\ + P(A \cap B \cap C) \end{Bmatrix}$

$P(A \cup B \cup C)$

. . $=\qquad\qquad\qquad\qquad P\bigg[(A \cup B) \cup C\bigg]$

. . $=\qquad\qquad P(A \cup B) + P(C) - P\bigg[ (A \cup B) \cap C\bigg]$

. . $=\qquad\qquad P(A \cup B) + P(C) - \underbrace{P\bigg[(A \cap C) \cup (B \cap C)\bigg]}$

. . $=\qquad P(A \cup B) + P(C) - \overbrace{\bigg[P(A\cap C) + P(B\cap C) - P([A \cap C] \cap [B \cap C])\bigg]}$

. . $=\qquad \underbrace{P(A \cup B)} + P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

. . $=\;\overbrace{P(A) + P(B) - P(A \cap B)} + P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

. . $=\; P(A) + P(B) + P(C) - P(A\cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

. . . . . . . . . . . . ta-DAA!