Mathematics 9709 · Probability & Statistics 1

Dependent events & trees

Dependent events mean the first thing changes the chance of the second.
The usual exam version is without replacement, or a tree where the second stage depends on the first.

The tree remembers what already happened

The general rule is $P(A\cap B)=P(A)P(B\mid A)$ . Multiply along a real path, using the probability that belongs to that path.

With replacement: the bag resets. The second probability stays the same.
Without replacement: one item has gone. The second denominator drops.
If a coin, die, or first choice picks the bag, keep that first probability on every path.

Worked example

First rung: two draws without replacement

A bag has 3 red counters and 2 blue counters. Two counters are drawn without replacement. Find the probability of two reds.

First red: $\tfrac35$ .
One red has gone, so second red is $\tfrac24$ .
Multiply along the path: $\tfrac35\times\tfrac24=\tfrac3{10}$ .

Answer

\dfrac3{10}

Head means one marble. Tail means two marbles without replacement, so the second tail branch has denominator 8.

Exam version: one tree, two denominators

A biased coin has $P(H)=\tfrac13$ . A bag has 4 red and 5 blue marbles. If H occurs, one marble is chosen. If T occurs, two marbles are chosen without replacement. (9709/52 M/J 2024 Q2)

At least one red after H: $\tfrac13\times\tfrac49=\tfrac{4}{27}$ .
At least one red after T: $\tfrac23\left(1-\tfrac59\times\tfrac48\right)=\tfrac{13}{27}$ .
Add the two working paths: $\tfrac{4}{27}+\tfrac{13}{27}=\tfrac{17}{27}$ .

Answer

P(\text{at least one red})=\tfrac{17}{27}

For given that on a tree, build the denominator from paths.

List every path where the condition happens.
Add those paths. That is the denominator.
Keep only the paths that also hit the target. That is the numerator.
Divide numerator by denominator.

In a tree question, $P(\text{target}\mid\text{condition})$ is not one branch over one branch. It is target-and-condition paths over all condition paths.

Common mistake

Do not reuse the first denominator after a selection without replacement. From 4 red and 5 blue, if a blue has gone, the next blue chance is

\tfrac48

, not

\tfrac59

The Bayes shape, without the scary name

The numerator is one route, or a few routes, that give both the target and the condition.
The denominator is every route that gives the condition.
This is why mark schemes reward a correct denominator even before the final answer.

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Dependent events & trees

Dependent events mean the first thing changes the chance of the second.
The usual exam version is without replacement, or a tree where the second stage depends on the first.

The tree remembers what already happened

The general rule is $P(A\cap B)=P(A)P(B\mid A)$ . Multiply along a real path, using the probability that belongs to that path.

With replacement: the bag resets. The second probability stays the same.
Without replacement: one item has gone. The second denominator drops.
If a coin, die, or first choice picks the bag, keep that first probability on every path.

Worked example

First rung: two draws without replacement

A bag has 3 red counters and 2 blue counters. Two counters are drawn without replacement. Find the probability of two reds.

First red: $\tfrac35$ .
One red has gone, so second red is $\tfrac24$ .
Multiply along the path: $\tfrac35\times\tfrac24=\tfrac3{10}$ .

Answer

\dfrac3{10}

Head means one marble. Tail means two marbles without replacement, so the second tail branch has denominator 8.

Exam version: one tree, two denominators

A biased coin has $P(H)=\tfrac13$ . A bag has 4 red and 5 blue marbles. If H occurs, one marble is chosen. If T occurs, two marbles are chosen without replacement. (9709/52 M/J 2024 Q2)

At least one red after H: $\tfrac13\times\tfrac49=\tfrac{4}{27}$ .
At least one red after T: $\tfrac23\left(1-\tfrac59\times\tfrac48\right)=\tfrac{13}{27}$ .
Add the two working paths: $\tfrac{4}{27}+\tfrac{13}{27}=\tfrac{17}{27}$ .

Answer

P(\text{at least one red})=\tfrac{17}{27}

For given that on a tree, build the denominator from paths.

List every path where the condition happens.
Add those paths. That is the denominator.
Keep only the paths that also hit the target. That is the numerator.
Divide numerator by denominator.

In a tree question, $P(\text{target}\mid\text{condition})$ is not one branch over one branch. It is target-and-condition paths over all condition paths.

Common mistake

Do not reuse the first denominator after a selection without replacement. From 4 red and 5 blue, if a blue has gone, the next blue chance is

\tfrac48

, not

\tfrac59

The Bayes shape, without the scary name

The numerator is one route, or a few routes, that give both the target and the condition.
The denominator is every route that gives the condition.
This is why mark schemes reward a correct denominator even before the final answer.