Conditional probability from a two-way table

What you will be able to do

• read a conditional probability straight from a two-way table;
• use $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ when the table gives shares instead of counts;
• explain why $P(A \mid B)$ and $P(B \mid A)$ are usually different.

A table from a survey

A made-up survey asks $200$ students two questions. Do you play an instrument? Are you in the school choir? A two-way table counts the students by both answers at once.

Pick a student at random. The chance that the student plays an instrument is $\dfrac{60}{200} = 0.3$.

Now suppose you know that the student sings in the choir. You can forget everyone outside the choir column. Of the $40$ choir students, $24$ play an instrument. So the chance is now

The bar is read as given. A probability that uses some known information is a conditional probability. Knowing "choir" doubled the chance of "instrument", from $0.3$ to $0.6$.

The other way round

What about $P(\text{choir} \mid \text{instrument})$? Now the known fact is "instrument", so look only along the instrument row. Of its $60$ students, $24$ are in the choir:

The top is the same, $24$. The bottom is not, because the group you look inside is different. In general $P(A \mid B)$ and $P(B \mid A)$ are different numbers. Mixing them up is one of the most common mistakes with probability, in the news as well as in class.

The rule behind the table

Divide the top and the bottom of $\dfrac{24}{40}$ by the total, $200$:

Here $\dfrac{24/200}{40/200} = \dfrac{0.12}{0.2} = 0.6$. The rule is useful when a report gives shares, not counts. You meet the exam form of this rule, and Bayes' theorem, in the Maths courses on conditional probability.

Try it in Python

A two-way table fits in a list of two lists: one list for each row. The program finds the totals, then the two conditional probabilities.

Your turn

The usual mistakes

• Dividing by the grand total. $P(A \mid B)$ divides by the count of $B$, not by everyone.
• Reading the wrong line. Given a column, read down that column. Given a row, read along that row.
• Swapping the two events. "Instrument given choir" and "choir given instrument" answer different questions.
• Confusing "and" with "given". $P(A \cap B)$ is a share of everyone. $P(A \mid B)$ is a share of $B$ only.

Remember these

• $P(A \mid B)$: look only at the row or column of $B$, and find the share of $A$ inside it.
• $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.
• $P(A \mid B)$ and $P(B \mid A)$ have the same top and different bottoms.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In the choir table, find P(no instrument given choir).

2. In the same table, find P(instrument given not choir).

3. A report says that $15\%$ of people are in a club and $6\%$ of people are in the club and are over $60$ years old. Find the probability that a club member is over $60$.

4. In a made-up survey, $P(\text{reads news} \mid \text{has a phone}) = 0.5$. Does that tell you $P(\text{has a phone} \mid \text{reads news})$?

5. Why do the conditional probabilities inside one column always add up to $1$?

References

• OpenStax, Introductory Statistics, Chapter 3, the section on contingency tables.