When Knowing One Thing Changes Another

A survey finds that 60% of residents use public transit. Of transit users, 30% commute before 7 a.m.; of nonusers, only 10% do. What fraction of all residents commute that early?

The answer is not 30% and not 10%. We must weight both subgroups.

Conditional probabilities live inside denominators

If $T$ denotes transit use and $E$ denotes early commuting, then

$$P(E\mid T)=\frac{P(E\cap T)}{P(T)}.$$

It means among transit users, what fraction commute early? Changing what we condition on changes the denominator and therefore the question.

Weighted branches

Partition the population by transit use:

$$P(E)=P(E\mid T)P(T)+P(E\mid T^c)P(T^c).$$

Using the fictional percentages gives

$$P(E)=.30(.60)+.10(.40)=.22.$$

So 22% commute early. A probability tree makes the same calculation visible: multiply along branches, then add disjoint branches ending in the event we care about.

This is called the law of total probability.

Check independence

Two events $A$ and $B$ are independent when

$$P(A\mid B)=P(A)$$

when conditioning is defined. Equivalently, $P(A\cap B)=P(A)P(B)$.

In our example, $P(E\mid T)=.30$ but $P(E)=.22$. The events are not independent. Knowing transit use changes the probability of early commuting.

Compare this with mutually exclusive events, which cannot happen together. A person cannot be simultaneously classified in two nonoverlapping transit categories, but the category and an early commute can certainly co-occur.

Why this matters for Bayesian reasoning

Bayes’ theorem reverses a conditional question by using the probability of the evidence across possible explanations. The base-rate laboratory is a probability tree in disguise: disease/no disease branches, followed by positive/negative test branches.

Always label which conditional probability each percentage represents. Sensitivity is not positive predictive value.

Check yourself

1. Find the probability that a resident both uses transit and commutes early.
2. If everyone commuted early at the same rate regardless of transit, what equality would hold?
3. Why can you add $P(E\cap T)$ and $P(E\cap T^c)$ without subtracting an overlap?

Worked responses

1. $.60(.30)=.18$.

2. $P(E\mid T)=P(E\mid T^c)=P(E)$, assuming both groups have positive probability.

3. Transit use and nonuse are disjoint events that partition the population.

Build the entire joint table

For 1,000 hypothetical residents, the stated percentages correspond to the following expected counts:

	Early	Not early	Total
Transit	180	420	600
No transit	40	360	400
Total	220	780	1000

Thus $P(T\mid E)=180/220=9/11\approx0.81818$, whereas $P(E\mid T)=180/600=0.30$. Conditioning in the other direction changes the denominator. The joint probability $0.18$ is the same intersection whichever event is named first, but the conditional probabilities are not interchangeable.

For a partition $B_1,\ldots,B_k$ with positive probabilities, $P(A)=\sum_iP(A\mid B_i)P(B_i)$. This follows by splitting $A$ into disjoint intersections $A\cap B_i$ and applying the multiplication rule. Bayes then divides one of those intersections by their total: $P(B_j\mid A)=P(A\mid B_j)P(B_j)/P(A)$ when $P(A)>0$.

Pairwise independence is weaker than mutual independence for three or more events. Toss two fair independent coins. Let A mean the first is heads, B the second is heads, and C mean the two agree. Each pair is independent, yet $P(A\cap B\cap C)=1/4$, not $(1/2)^3$. Conditioning on two events can reveal a dependence invisible in pairwise checks.

Further problems

1. From the table, compute $P(T\mid E^c)$.
2. If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.2$, are A and B independent? Mutually exclusive?
3. Why is conditioning on an event of probability zero not defined by the elementary ratio formula?

Worked further solutions

1. $420/780=7/13\approx0.53846$.
2. They are independent because the intersection equals the product. They are not mutually exclusive because the intersection probability is positive.
3. The denominator would be zero. Conditional distributions at particular values of continuous variables require more advanced definitions than this elementary event formula.