Conditioning and the chain rule

Idea: News crosses out outcomes

You learned something, so some outcomes are gone. Work on the list that is left.

• The survivors stay equally likely if the news favours none
• Count the event inside the survivors

Example: First die 6, given a total of 9 or more

Roll $2$ dice. The total is at least $9$. What is the chance the first die shows $6$?

1. Totals $9$ to $12$: $4+3+2+1=10$ pairs remain
2. Starting with $6$: $(6,3)$, $(6,4)$, $(6,5)$, $(6,6)$
3. Chance: $\frac{4}{10}=\frac{2}{5}$

Formula: Conditional probability

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

The share of B that is also A. It works even when outcomes are not equally likely.

Formula: The chain rule

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

Each factor is a chance on the list left after the earlier steps.

Example: Both cards hearts

Draw $2$ cards from a $52$-card deck without replacement. What is the chance both are hearts?

1. First: $\frac{13}{52}$
2. Second, with a heart gone: $\frac{12}{51}$
3. Both: $\frac{13}{52}\times \frac{12}{51}=\frac{1}{17}$

Idea: Moves in a chain

Chains of draws combine with the usual tools.

• Exclusive paths add: $\frac{5}{8}\times \frac{3}{7}+\frac{3}{8}\times \frac{5}{7}=\frac{15}{28}$
• A free first draw: a matching suit needs $\frac{12}{51}$
• Complements work: $1-\frac{39}{52}\times \frac{38}{51}=\frac{15}{34}$

Common mistake: Which die is the news about?

"The first die shows $6$" and "at least one die shows $6$" are different news.

Instead The first leaves $6$ outcomes, so both dice show $6$ with $\frac{1}{6}$. The second leaves $36-25=11$, so $\frac{1}{11}$.

Quick check: Pick an urn, see red

Urn $A$: $3$ red, $1$ blue. Urn $B$: $1$ red, $1$ blue. A fair coin picks an urn and you draw red. Chance it was $A$?

Idea: Say it out loud

State what you know, then the size of the list that remains.

• In a chain, say "given the first draw" for each factor
• Say what changed in the deck
• Ask whether the way you heard the news depends on the outcome