Conditional Probability (Part 1)
Last modified — 02 Oct 2026
By the end of this lecture, students are anticipated to be able to:
In general, the outcome of a random experiment can be any element of \(\Omega\).
Sometimes, we have “partial information” about which elements can occur.
Roll a die.
If \(A\) is the event of obtaining a “2”, then \(\mathbb{P}(A) = 1/6\).
But if the outcome is known to be even, then intuition suggests that \(\mathbb{P}(A) > 1/6\).
Two events play distinct roles in this example:
The event of interest \(A = \{ 2 \}\)
The conditioning event \[ B= \{\text{outcome is even}\} = \{2, 4, 6\} \]
The conditioning event captures the “partial information”
If we only consider the three possible outcomes in B (even numbers), only one of them is a 2. Therefore, the probability of rolling a 2 if you know the roll was even is indeed 1/3.
The probability of an event A, conditional on event B is written as
\[\mathbb{P}(A \ \vert\ B)\]
We read this as “the probability of \(A\) given \(B\)” or “the probability of \(A\) conditional on \(B\)”
Just as \(\mathbb{P}(\cdot)\) is a function, for any fixed \(B\), \(P \left(\ \cdot \ \ \vert\ B \right)\) is also a function. Its argument is any event \(A \subseteq \Omega\).
Moreover, \(\mathbb{P}\left(\ \cdot \ \ \vert\ B \right)\) satisfies the three Axioms of a Probability (i.e., is a probability).
100 people were surveyed about their pets. 20 people had a cat, and 12 people had both a dog and a cat. What is the probability that a randomly selected person has a dog, given that they have a cat?
As \(P(\cdot \ \vert\ B)\) is a probability, for any event \(A\):
\[\mathbb{P}( A^C \ \vert\ B) = 1 - \mathbb{P}(A \ \vert\ B)\]
Your friend flips two coins, looks at it, and tells you that the two faces are the same.
What is the probability that both coins show heads? Use the definition of conditional probability to solve this.
If \(\mathbb{P}(A_1) > 0\): \[ \mathbb{P}\left( A_{1} \cap A_{2}\right) = \mathbb{P}\left(A_{2}\ \vert\ A_{1}\right) \, \mathbb{P}\left( A_{1}\right). \] This is just a rearrangement of our definition!
If \(\mathbb{P}(A_1),\ \mathbb{P}(A_1 \cap A_2),\dots,\ P(A_1 \cap A_2 \cap \dots \cap A_{n-1}) > 0\), then \[\begin{aligned} \mathbb{P}\left( A_{1}\cap A_{2}\cap \cdots \cap A_{n}\right) &= \mathbb{P}\left( A_{n}\ \vert\ A_{1}\cap A_{2}\cap \cdots \cap A_{n-1}\right) \\ &\quad \times \mathbb{P}\left( A_{n-1}\ \vert\ A_{1}\cap A_{2}\cap \cdots \cap A_{n-2}\right) \\ & \quad \times \cdots \times\\ & \quad \times \mathbb{P}\left( A_{3}\ \vert\ A_{1}\cap A_{2}\right) \times \mathbb{P}\left( A_{2}\ \vert\ A_{1}\right) \times \mathbb{P}\left( A_{1}\right) \end{aligned}\]
An urn has 10 red balls and 40 black balls.
Three balls are randomly drawn without replacement.
Calculate the probability that:
The 3rd ball is red given that the 1st is red and the 2nd is black.
The first drawn ball is red, the 2nd is black and the 3rd is red.
We say that \(B_{1}, \ldots, B_{n}\) is a partition of \(\Omega\) if
They are disjoint \[ B_{i}\cap B_{j} \, = \, \varnothing \quad \mbox{ for } i \ne j \, , \]
They cover the whole sample space: \(\bigcup_{i=1}^{n} B_{i} \, = \, \Omega\)
A simple partition is any event \(A\) and its complement \(A^c\).
If \(B_{1}, \ldots, B_{n}\) is a partition of \(\Omega\), then, for any \(A \subset \Omega\), \[\mathbb{P}\left( A\right) =\sum_{i=1}^{n} \mathbb{P}\left( A\ \vert\ B_{i}\right) \, \mathbb{P}\left( B_{i}\right).\]
\(A = A \cap \Omega = A \cap \left( \bigcup _{i=1}^{n}B_{i}\right) = \bigcup_{i=1}^{n}\left( A \cap B_{i}\right)\)
The events \(\left( A \cap B_{i}\right)\) are disjoint.
Therefore, by Axiom 3, we have \[\begin{aligned} \mathbb{P}\left( A\right) & = \mathbb{P}\left( \bigcup_{i=1}^{n} A \cap B_{i} \right) \\ &=\sum_{i=1}^{n} \mathbb{P}\left( A\cap B_{i}\right) \\ &=\sum_{i=1}^{n} \mathbb{P}\left( A\ \vert\ B_{i}\right) \, \mathbb{P}\left( B_{i}\right). \end{aligned}\]
If a new patient walks into the ER, what is the probability that they test positive for the flu?
Your first checkpoint (one short question at the end of class ~ 15 minutes) is on Friday September 25th
It is a closed-book assessment based on the practice problems assigned for Sections 1.1 - 1.4 (all material so far up to the end of this lecture)
You may bring a non-graphing, non-programmable calculator and one cheat sheet, which you should add/edit to over time for all assessments. The cheat sheet must be:
Cheat sheets and calculators that do not follow these rules will be confiscated - sorry!
Stat 302 - Winter 2025/26