Lecture 4

Conditional Probability

Conditional Probability: The conditional probability that \(A\) has occurred given that \(B\) has occurred with \(P(B) \geq 0\) is defined as \[ P(A|B) = \frac{P(A\cap B)}{P(B)} \]

Rather than thinking about \(P(A|B)\) within universe \(\Omega\), think of \(B\) as replacing \(\Omega\).

\(A\) can be thought of as an event that occurs within the universe \(B\).

If the possible outcomes are finitely many and equally likely, then \[ P(A|B) = \frac{\text{number of elements in }A\cap B}{\text{number of elements in }B} \]

Example: A bin contains 25 light bulbs, 5 of which are in good condition (last at least a month), 10 of which are partially defective (works initially for 2 days), and 10 which are totally defective. Given that a bulb is lit initially for 2 days, what is the probability that it will still be working after a week?

  • Because the bulb works initially for 2 days, we know it's not totally defective. We can consider this \(D^c\), or the complement of the defective set. \(D^c=\frac{15}{25}\) because \(\frac{10}{25}\) are totally defective.

  • We're looking for the probability that a bulb is perfectly good (\(G\)) given that it's not totally defective. The intersection of the non-defective bulbs and the perfectly good bulbs is \(\frac{5}{25}\) - in other words, 5 of the remaining bulbs are in good condition.

\[ P(G|D^c) = \frac{G\cap D^c}{P(D^c)} = \frac{\frac{5}{25}}{\frac{15}{25}}=\frac{1}{3} \]

The Law of Total Probability

Let \(A\) and \(B\) be events. We may express \(A\) as \[ A = (A \cap B) \cup (A \cap B^c) \]

We can also say that \[ \begin{align} P(A) &= P(AB) + P(AB^c)\\ &=P(A|B)P(B) + P(A|B^c)P(B^c) \end{align} \]

If the sample space \(\Omega\) can be partitioned into \(n\) mutually exclusive events \(A_1, A_2, \dots, A_n\), then \[ \begin{align} P(A) &= P(A \cap B_1) + P(A \cap B_2) + \dots + P(A \cap B_n)\\ &=P(A|B_1)P(B_1) + \dots + P(A|B_n)P(B_n) \end{align} \]

Bayes' Formula

We can find the conditional probability \(P(A_i|B)\), in terms of \(P(B|A)\) using the law of total probability and Bayes' rule.

Let \(A_1, A_2, \dots, A_n\) be disjoint sets that form a partition of the sample space, and assume that \(P(A_i) > 0\) for all \(i\). Then, for any event \(B | P(B) > 0\), we have \[ \begin{align} P(A_i|B) &= \frac{P(A_i)P(B|A_i)}{P(B)}\\ &=\frac{P(A_i)P(B|A_i)}{P(A_1)P(B|A_1) + \dots + P(A_n)P(B|A_n)} \end{align} \]

Example: A bowl contains three apples and four oranges. Bob walks by and randomly selects a fruit. Steve walks by 10 minutes later and also randomly selects a fruit from the bowl.

Let \(B_a\) be the event that Bob choses an apple, \(B_o\) the event that Bob chooses an orange, \(S_a\) the event that Steve chooses an apple, and \(S_o\) the event that Steve chooses an orange.

What is the probability \(P(S_a)\) that Steve selects an apple? Use the law of total probability.

\[ \begin{align} P(B_a) &= \tfrac{3}{7}\\ P(B_o) &= \tfrac{4}{7}\\ P(S_a | B_a) &= \tfrac{2}{6} = \tfrac{1}{3}\\ P(S_a | B_o) &= \tfrac{3}{6} = \tfrac{1}{2}\\ P(S_a) &= P(S_a | B_a)P(B_a) + P(S_a | B_o)P(B_o)\\ &= \tfrac{1}{3}\cdot\tfrac{3}{7} + \tfrac{1}{2}\cdot\tfrac{4}{7} = \tfrac{3}{7} \end{align} \]

If Steve selected an orange, what's the possibility that Bob had also selected an orange?

\[ P(B_o | S_o) = \frac{}{} = \frac{\left(\tfrac{1}{2}\right)\left(\tfrac{4}{7}\right)}{\left(\tfrac{1}{2}\right)\left(\tfrac{4}{7}\right) + \left(\tfrac{2}{3}\right)\left(\tfrac{3}{7}\right)} = \frac{1}{2} \]

Independent Events: Events \(A\) and \(B\) are said to be independent if

\[ \begin{align} P(A | B) &= \tfrac{P(AB)}{P(B)} = P(A)\text{, or}\\ P(AB)&=P(A)P(B) \end{align} \]