Theorem 2.2

#Theorem

Let $A$ and $B$ be events. Then $P (A \cup B) = P (A) + P (B) - P (A \cap B)$

If $A$ and $B$ are mutually exclusive events, then $P (A \cup B) = P (A) + P (B)$

Proof:

Let $A = {1, 2, 3, 4, 5}$
Let $B = {4, 5, 6, 7}$

$∴ A \cup B = {1, 2, 3, 4, 5, 6, 7}$
$A \cap B = {4, 5}$
However, $A + B = {1, 2, 3, 4, 4, 5, 5, 6, 7}$
$⟹ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Example

A fair coin is tossed 3 times, and a sequence of heads and tails is observed. What is the probability that an even number of heads or at least two tails appear?

$S =$ The set of sequences of length 3 whose elements are from the set ${H, T}$
So $| S | = 2^{3} = 8$
$A =$ The event that an even number of heads appears $= {H H T, H T H, T H H, T T T}$
$B =$ The event that at least two tails appear $= {T T H, T H T, H T T, T T T}$
$A \cap B =$ The event that an even number of heads and at least two tails appear $= {T T T}$
$P (A) = \frac{4}{8} = \frac{1}{2}, P (B) = \frac{4}{8} = \frac{1}{2}, P (A \cap B) = \frac{1}{8}$
$⟹ P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{8} = \frac{7}{8}$