The conditional probability formula

We've seen that when outcomes are equally likely, we find $P(A \mid B)$ by restricting to the known event $B$ and counting. Counting relies on every outcome being equally likely. When outcomes have different probabilities, we restrict to $B$ in the same way, but measure each region by its probability rather than by its number of outcomes.

The diagram below conditions on $B$ in steps, with the probability of each region written in.

Restricting and rescaling work the same way for any events, so we take the result as the definition of conditional probability.

We need $P(B) > 0$ because the formula divides by $P(B)$: an event that cannot happen can never be learned, so there is nothing to restrict to.

A die is loaded: faces $1$ to $6$ have probabilities $0.1$, $0.3$, $0.2$, $0.1$, $0.1$ and $0.2$. Given that the roll is at most $3$, find the probability that it is odd.

Solution

We name the two events. The roll being at most $3$ is what we know, so it is the known event:

• $B = \{1, 2, 3\}$, the roll is at most $3$.
• $A = \{1, 3, 5\}$, the roll is odd.

We want $P(A \mid B)$. The table marks the faces in $B$ and those also in $A$:

Adding the marked probabilities:

We rescale by dividing by $P(B)$:

Given that the roll is at most $3$, it is odd with probability $\dfrac{1}{2}$.

Counting faces would give $\dfrac{2}{3}$, since two of the three faces in $B$ are odd. That treats faces $1$, $2$ and $3$ as equally likely, but face $2$ alone has probability $0.3$, which is half of $P(B)$.

A probability function is any function that satisfies the two axioms of probability. If conditional probabilities given $B$ satisfy them as well, they are probabilities, and every rule that follows from the axioms holds for them too. We check each axiom in turn, using the conditional probability formula.

Axiom 1 says the impossible event has probability $0$ and the certain event has probability $1$. Given $B$, the impossible event still has no outcomes, so $\emptyset \cap B = \emptyset$ and

Once we know that $B$ happens, every outcome left is in $B$, so $\Omega \cap B = B$ and

So the probabilities given $B$ total $1$, and axiom 1 holds.

Axiom 2, additivity, holds given $B$ as well. If two events $A_1$ and $A_2$ have no outcome in common, neither do their parts inside $B$, and rescaling divides every probability by the same number, $P(B)$, so a sum of probabilities stays a sum:

This is additivity with $\mid B$ written after every probability.

So conditional probabilities given $B$ satisfy both axioms: they are probabilities. Every rule that follows from the axioms, such as the complement rule and the general addition rule, therefore holds given $B$.

In a sports club, $40\%$ of members play tennis and $15\%$ play both tennis and squash. A member chosen at random plays tennis. Find the probability that they do not play squash.

Solution

We name the events. That the member plays tennis is what we know, so it is the known event:

• $T$: the member plays tennis, with $P(T) = 0.4$.
• $S$: the member plays squash, with $P(S \cap T) = 0.15$.

We want $P(S^c \mid T)$. The complement rule holds given $T$:

So we first find $P(S \mid T)$ with the conditional probability formula:

Substituting into the complement rule:

A member who plays tennis does not play squash with probability $0.625$.