# Probability Theory

Daniel Weibel
Created 8 Nov 2017
$\newcommand{\P}{\text{P}} \newcommand{\Vert}{\,\vert\,}$

# Illustration # Conditional Probability

## Probability of B Given A

$\P(B\Vert A)$
• If $A$ occured, what is the probability that now also $B$ occurs?
• Percentage of $A$ that is also in $B$.

## Probability of A Given B

$\P(A\Vert B)$
• If $B$ occured, what is the probability that now also $A$ occurs?
• Percentage of $B$ that is also in $A$.

# Probability of A And B

\begin{align*} \P(A \cap B) & = \P(A) \cdot \P(B\Vert A) \\ & = \P(B) \cdot \P(A\Vert B) \end{align*}

# Probability of A Or B

$\P(A \cup B) = \P(A) + \P(B) - \P(A \cap B)$

# Conditional Probability Revisited I

## Probability of B Given A (I)

$\P(B\Vert A) = \frac{\P(A \cap B)}{\P(A)}$

Deduced from this formula for $\P(A \cap B)$.

## Probability of A Given B (I)

$\P(A\Vert B) = \frac{\P(A \cap B)}{\P(B)}$

Deduced from this formula for $\P(A \cap B)$.

# Conditional Probability Revisited II

## Probability of B Given A (II)

$\P(B\Vert A) = \frac{\P(B) \cdot \P(A\Vert B)}{\P(A)}$

In this formula for $\P(B\Vert A)$, replace $\P(A \cap B)$ with this formula for $\P(A \cap B)$.

This equation is called Bayes’ Theorem.

## Probability of A Given B (II)

$\P(A\Vert B) = \frac{\P(A) \cdot \P(B\Vert A)}{\P(B)}$

In this formula for $\P(A\Vert B)$ replace $\P(A \cap B)$ with this formula for $\P(A \cap B)$.

This equation is called Bayes’ Theorem.

# Independent Events

The happening of one event has no effect on the probability of the other event.

For example:

• $A$ = getting head on first toss of a coin
• $B$ = getting head on second toss of a coin
$\P(A \cap B) = \P(A) \cdot \P(B)$

In this formula for $\P(A \cap B)$, replace $\P(B\Vert A)$ with $\P(B)$.

We can do this, because if $A$ occurred, the probability that $B$ occurs is not affected by that. The probability that $B$ occurs is still $\P(B)$.

# Mutual Exclusive Events

The happening of one event prevents the happening of the other event.

For example:

• $A$ = rolling a 1 with a die
• $B$ = rolling a 2 with a die
$\P(A \cup B) = \P(A) + \P(B)$

In this formula for $\P(A \cup B)$, replace $\P(A \cap B)$ with 0.

We can do this, because events $A$ and $B$ cannot occur together (the probability that they occur together is 0).