POLS 3220: How to Predict the Future
Last time, we defined probability as a number between 0 and 1 describing the likelihood of an event.
Today, our focus is compound probability, the likelihood of some combination of two or more events.
Compound probability is particularly counterintuitive.
By the end of the lecture, you’ll be equipped with some tools for dealing with these sorts of problems.
Imagine the future as a series of branching paths.
Each branch represents a different path that the universe could take.
Branches originating from a single point must be mutually exclusive and collectively exhaustive.
Mutually Exclusive: if one event happens, the others cannot happen
Collectively Exhaustive: covers every possible outcome
Think about a sequence of 2-3 choices you expect to make after class is over (e.g. where you’ll eat dinner). Draw a tree illustrating all the possible paths your evening could take.
Next, we will assign a probability to each branch. When doing so, remember two rules:
These two rules are called the axioms of probability.
The second set of probabilities are called conditional probabilities.
We denote the probability of event \(A\) conditional on event \(B\) as:
\[ P(A|B) \]
And the probability of \(A\) conditional on \(B\) not happening is:
\[ P(A|\neg B)\]
Two events are independent if the outcome of one doesn’t affect the probability of the other. Formally:
\[P(A|B) = P(A|\neg B)\]
Coin flips are a classic example of independent events:
Assign probabilities to each branch of your evening tree. Make sure you adhere to the axioms. And consider whether your choices are independent, or if one event might affect the probabilities of subsequent events.
Now we’re ready to tackle joint probability. What is the probability of event \(A\) and event \(B\) both happening?
To find the probability of ending up at any node of the tree, multiply the probabilities of all the branches that feed into it.
Calculate joint probabilities for every node in your evening tree. To check your work, make sure everything satisfies the axioms of probability.
This gives us some insight into the birthday problem we started with: