The multiplication rule and tree diagrams

What you will be able to do

• use $P(A \cap B) = P(A)\,P(B \mid A)$ to find the chance that two things both happen;
• draw a tree diagram, multiply along the branches and add the paths that make an event;
• find "at least one" with the complement.

Two cards from a pack

Two cards are taken from a pack of $52$, one after the other, and the first is not put back. What is the chance that both are aces?

The first card is an ace with probability $\dfrac{4}{52}$. Now $3$ aces are left among $51$ cards. So the second card is an ace, given that the first was, with probability $\dfrac{3}{51}$. Both happen with probability

We multiplied a first-stage probability by a second-stage probability that already knew what happened first. That is the multiplication rule.

The multiplication rule

Start from the definition $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$ and multiply both sides by $P(A)$:

By symmetry $P(A \cap B) = P(B)\,P(A \mid B)$ too. Indian books call this the multiplication theorem. For three events the rule goes on:

Each factor is conditional on everything that came before it.

Tree diagrams

A tree diagram shows an experiment stage by stage. Each stage is a set of branches. Write on each branch the probability of that step, given the path so far.

• The branches leaving one point add up to $1$, because exactly one of them must happen.
• Multiply along a path to get the probability of that whole outcome.
• Add the paths that belong to an event.

The tree below is for a bag with $4$ red and $3$ blue balls, drawn twice without replacement. After a red first draw, the bag has $3$ red and $3$ blue balls, so the second-stage branches are $\dfrac{3}{6}$ and $\dfrac{3}{6}$.

With and without replacement

If the first ball is put back, the bag is the same for the second draw, and the second-stage branches repeat the first. Then $P(\text{RR}) = \dfrac{4}{7} \times \dfrac{4}{7} = \dfrac{16}{49}$. Without replacement, $P(\text{RR}) = \dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{2}{7} = \dfrac{14}{49}$. The branches of a tree change only when the first stage changes the second.

A worked example with three stages

A box holds $6$ good bulbs and $4$ faulty bulbs. Three bulbs are tested one by one without replacement.

1. All good. $\dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{1}{6}$.
2. At least one faulty. Use the complement: $1 - \dfrac{1}{6} = \dfrac{5}{6}$.
3. Exactly one faulty. There are three paths: FGG, GFG and GGF. The path FGG has probability $\dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{1}{6}$. The other two paths have the same three numerators in another order, so they are $\dfrac{1}{6}$ each. The sum is $\dfrac{1}{2}$.

The shortcut in step 2 is worth remembering: $P(\text{at least one}) = 1 - P(\text{none})$. It replaces many paths with one.

Practise with new numbers

Try it in Python

The program walks through the tree of the bag with $4$ red and $3$ blue balls. It multiplies along each path and adds the paths that give exactly one red.

Common mistakes

• Keeping the first-stage probability for the second stage. Without replacement the numbers in the bag have changed.
• Adding along a path. On a path you multiply. You add only across different paths.
• Missing a path. For "exactly one red" there are two paths, red then blue and blue then red. Count the orders.
• Forgetting the complement. For "at least one" it is faster to find "none" and subtract from $1$.

Remember these

• $P(A \cap B) = P(A)\,P(B \mid A)$, and for three events $P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)$.
• In a tree, multiply along a path and add across paths.
• The branches from one point add up to $1$.
• $P(\text{at least one}) = 1 - P(\text{none})$.

Check yourself

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

1. Two cards are drawn from a pack of $52$ without replacement. Find the probability that both are kings.

2. A bag holds $5$ red and $3$ blue balls. Two balls are drawn with replacement. Find the probability that both are red, and the probability that the two colours differ.

3. The same bag of $5$ red and $3$ blue balls is used, but now the first ball is not put back. Find the probability that exactly one ball is red.

4. A fair die is thrown $4$ times. Find the probability of at least one six.

5. Why must the branches leaving one point of a tree add up to $1$?

References

• OpenStax, Introductory Statistics, Chapter 3, section on tree and Venn diagrams.
• NCERT, Mathematics Part II for Class XII, Chapter 13, section on the multiplication theorem on probability.