Conditional Probability

138. Conditional Probability

Probability of $𝐴$ given that $𝐵$ has occurred

𝑃 (𝐴 | 𝐵) = \frac{𝑃 (𝐴 \cap 𝐵)}{𝑃 (𝐵)}

Note:

𝑃 (𝐴 | 𝐵) \geq 0

𝑃 (𝐴 | 𝐴) = 1

𝑃 (Ω | 𝐴) = 1

Additivity:

\begin{matrix} 𝑃 (𝐴 \cup 𝐶 | 𝐵) = 𝑃 (𝐴 | 𝐵) + 𝑃 (𝐶 | 𝐵) \\ if 𝐴 \cap 𝐶 = \emptyset \end{matrix}

\begin{aligned} \frac{𝑃 ((𝐴 \cup 𝐶) \cap 𝐵)}{𝑃 (𝐵)} & = \frac{𝑃 ((𝐴 \cap 𝐵) \cup (𝐶 \cap 𝐵))}{𝑃 (𝐵)} \\ = \frac{𝑃 (𝐴 \cap 𝐵) + 𝑃 (𝐶 \cap 𝐵)}{𝑃 (𝐵)} \\ = 𝑃 (𝐴 | 𝐵) + 𝑃 (𝐶 | 𝐵) \end{aligned}

Example

Event $𝐴$ : Airplane is flying above

Event $𝐵$ : Something registers on radar screen

\begin{aligned} 𝑃 (𝐴) & = 0.05 \\ 𝑃 (𝐴^{𝑐}) & = 0.95 \end{aligned}

𝑃 (𝐴 \cap 𝐵) = 𝑃 (𝐴) \cdot 𝑃 (𝐵 | 𝐴) = 0.05 \cdot 0.99

𝑃 (𝐵) = 0.05 \cdot 0.99 + 0.95 \cdot 0.10

\begin{aligned} 𝑃 (𝐴 | 𝐵) & = \frac{𝑃 (𝐴 \cap 𝐵)}{𝑃 (𝐵)} \\ = \frac{0.05 \cdot 0.99}{0.05 \cdot 0.99 + 0.95 \cdot 0.10} \end{aligned}