Lecture 3

Uniform Probability on Finite Spaces and Conditional Probabilities

Grace Tompkins

Last modified — 23 May 2026

Last Class:

Defined probability as a mathematical object
Used set-theory to compute probabilities and prove probabilistic properties

Today’s Learning Outcomes

By the end of this lecture, students are anticipated to be able to

Calculate the number of possible outcomes using:
- Equally likely outcomes
- Combinations
Define a conditional probability.

We will start with some word problems from last class’ material

1 Applied Problems

Applied Problems

Marley borrows 2 books. Suppose that there is a 0.5 probability they like the first book, 0.4 that they like the second book, and 0.3 that they like both.

What is the probability that they will NOT like both books? (i.e. that they will not like either book?)

Applied Problems

Jane must take two tests, call them \(T_1\) and \(T_2\). The probability that she passes test \(T_1\) is 0.8, that she passes test \(T_2\) is 0.7, and that of passing both tests is 0.6.

Calculate the probability that:

She passes at least one test.
She passes at most one test.
She fails both tests.
She passes only one test.

Applied Problems

Suppose that \(\mathbb{P}\left( A\right) =0.85\) and \(\mathbb{P}\left( B\right) =0.75.\) Show that \[\mathbb{P}\left( A\cap B\right) \geq 0.60.\]

Applied Problems

2 Uniform Probability on Finite Spaces

Finite and Equally Likely Outcomes

When there are finitely many outcomes, and they are equal likely, calculating probabilities involves counting outcomes in events/sets,

Let \(A\) be a subset (event) of a sample space \(\Omega\).

\[ \mathbb{P}\left( A\right) = \frac{\text{number of elements in }A}{\text{number of elements in }\Omega } \]

Finite and Equally Likely Outcomes

Experiment: roll a fair die;
Sample space: \(\Omega = \{1, 2, 3, 4, 5, 6\}\).
If the die is fair we have \[\mathbb{P}(\{1\}) = \mathbb{P}(\{2\}) = \cdots = \mathbb{P}(\{6\})\]

We have 6 possible outcomes, all equally likely.

Therefore the probability of rolling a any single number is \(1/6\).

Finite and Equally Likely Outcomes

Suppose that we flip three different fair coins. What is the probability of rolling three heads in a row?

Finite and Equally Likely Outcomes

When the number of possible events are small, solving problems in this way is straightforward.
However: counting the number possible events can be challenging, particularly as the number of possible events increases
- Imagine rolling flipping 10 coins in a row. Writing out the sample space would be unreasonablly long
We will introduce permutations and combinations briefly to overcome this issue
- These provide ways to count the number of possible events

Counting Sequences: Multiplicative Principle

If a random experiment has k steps.

Step 1 has \(n_1\) possible outcomes,
Step 2 has \(n_2\) possible outcomes,

\(\quad\quad\quad\vdots\quad\quad\quad\)

Step k has \(n_k\) possible outcomes.

Then, \[\mbox{total number of outcomes} \ = \ n_1 \times n_2 \times n_3 \times \cdots \times n_k\]

Note

Implicit assumption: the outcomes of each step do not depend on each other.

Counting Sequences: Multiplicative Principle

Suppose that we flip three different fair coins. Without writing out the sample space, can you calculate the probability of rolling a head, a tail, and then a head?

Counting Sequences: Multiplicative Principle

What is the probability of drawing 5 cards in a row that are all clubs? Assume any card you picked is put back into the deck and shuffled before every draw.

3 Combinations

Combinations

Combination (of size \(m\)): a subset of \(m\) items from a set of size \(n\) (where necessarily \(m\) \(\leq\) \(n\)).

Note: We only care which elements are in the set, not the ordering.

Consider the set \[S=\left\{ a, b, c, d, e\right\}\]
The following are all the possible subsets of \(S\) of size 3:

\(\{ a, b, c\}\)	\(\left\{ a, d, e\right\}\)
\(\{ a, b, d\}\)	\(\left\{ b, c, d\right\}\)
\(\{ a, b, e\}\)	\(\left\{ b, c, e\right\}\)
\(\{ a, c, d\}\)	\(\left\{ b, d, e\right\}\)
\(\{ a, c, e\}\)	\(\left\{ c, d, e\right\}\)

Combinations

Note that the order of the elements does not matter (these are sets, not sequences).

\[ \Bigl\{ a, b, d \Bigr\} \, = \, \Bigl\{ d, a, b \Bigr\} \, = \, \Bigl\{ b, d, a \Bigr\} \] (and other possible rearrangements).

Combinations

A useful mathematical property is the factorial.

Factorial (!):

For any non-negative integer \(n\):

\[ n! = n \times(n-1)\times(n-2)\times(n-3)...3\times2\times1 \] and

\[ 0! = 1 \]

Number of Combinations

The number of combinations of size \(m\) out of a set of size \(n \ge m\) has various notations:

\[\binom{n}{m} = \left._{n} C_{m}\right. = C_m^n = \frac{n!}{m!(n-m)!}\]

\(n\): size of the set from which combinations are drawn
\(m\): size of the combinations
we read it as “n choose m”

Tip

In this course we use \(\binom{n}{m}\).

Example

For example, when \(n=5\) and \(m=3\) we have

\(\{ 1,2,3\}\)	\(\left\{ 1,4,5\right\}\)
\(\{ 1,2,4\}\)	\(\left\{ 2,3,4\right\}\)
\(\{ 1,2,5\}\)	\(\left\{ 2,3,5\right\}\)
\(\{ 1,3,4\}\)	\(\left\{ 2,4,5\right\}\)
\(\{ 1,3,5\}\)	\(\left\{ 3,4,5\right\}\)

hence, the number of combinations must be

\[\binom{5}{3}= {10}\]

General formula

\[\binom{n}{m} = \frac{n!}{m!(n-m)!}\] are also called binomial coefficients.

They have many beautiful interpretations.

Lotto 6/49

Dealer secretly chooses 6 distinct integers between 0 and 49.
Player randomly selects 6 distinct integers between 0 and 49.
The more matches, the bigger the prize.

What is the probability that the player matches \(k\) numbers from the dealer’s selection (for different values of \(k \in \{0, 1, ..., 6 \}\))?

Lotto 6/49

4 Permutations

Permutations

A permutation of a set is an arrangement of its elements in a specific order.

For example, calculating the number of ways:

3 people can sit next to each other in 3 empty seats
10 people can be randomly assigned to three distinct volunteer positions

Suppose \(|\Omega|\) = n and we want to count the number of permutations of length \(k \le n\) obtained from \(\Omega\).

In general, there are \(n \times n - 1 \times ... \times n - k + 1\) permutations of length k from a set of n elements.

Permutations

How many ways can the letters ABCDEF be rearranged?

Permutations

In how many different ways can the letters of “pepper” be arranged?

5 Conditional Probability

Conditional Probability

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 \(\mathbb{P}(A) > 1/6\).

Conditional probability formalizes this intuition (and helps to avoid mistakes)

Conditional Probability

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.

Conditional Probability

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 \(B\)”

Conditional Probability (Formal definition)

Let \(A, B \subseteq \Omega\) and assume \(\mathbb{P}(B) > 0\)

The conditional probability of \(A\) given \(B\) is \[\mathbb{P}\left(A \ \vert\ B \right) \, = \, \frac{ \mathbb{P}\left( A \cap B \right) }{ \mathbb{P}\left( B \right) }\]

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).

Useful Result:

As \(P(\cdot \ \vert\ B)\) is a probability, for any event \(A\):

\(\mathbb{P}(A \ \vert\ B) + \mathbb{P}( A^C \ \vert\ B) = 1\)
This implies:

\[\mathbb{P}( A^C \ \vert\ B) = 1 - \mathbb{P}(A \ \vert\ B)\]

Formalizing our intuition

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.

Lecture 3

Grace Tompkins

Last Class:

Today’s Learning Outcomes

1 Applied Problems

Applied Problems

Applied Problems

Applied Problems

Applied Problems

Applied Problems

2 Uniform Probability on Finite Spaces

Finite and Equally Likely Outcomes

Finite and Equally Likely Outcomes

Finite and Equally Likely Outcomes

Finite and Equally Likely Outcomes

Counting Sequences: Multiplicative Principle

Counting Sequences: Multiplicative Principle

Counting Sequences: Multiplicative Principle

3 Combinations

Combinations

Combinations

Combinations

Number of Combinations

Example

General formula

Lotto 6/49

Lotto 6/49

4 Permutations

Permutations

Permutations

Permutations

5 Conditional Probability

Conditional Probability

Conditional Probability

Conditional Probability

Conditional Probability (Formal definition)

Useful Result:

Formalizing our intuition

Formalizing our intuition

To Do:

Next Week