Dependent events and probability trees | CAIE 9709 Probability & Statistics 1 Notes

Chapters

Probability & Statistics 1

Mathematics9709

Replacement decides whether the bag resets

With replacement: the first item returns. Counts and probabilities reset, so random repeated draws are independent.
Without replacement: one item has gone. The next denominator is one smaller and the relevant numerator may also change.
At every tree node, the outgoing branch probabilities must add to 1.

Worked example

Two reds without replacement

A bag contains 3 red and 2 blue counters.

Answer

P(RR)=\frac35\times\frac24=\frac3{10}

Common mistake

After drawing a red, the second red probability is

2/4

, not

3/5

. Both the total and the red count changed.

Multiply along a path; add alternative paths

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

A complete path joins events with AND, so multiply its branches.
Different successful paths are alternatives joined by OR, so add their path probabilities.
For “at least one”, the complement of no successes is often shorter.

Worked example

One red and one blue in either order

The successful paths are RB and BR.

P(RB)=\frac35\frac24=\frac3{10}

P(BR)=\frac25\frac34=\frac3{10}

Answer

\frac3{10}+\frac3{10}=\frac35

A tree condition is built from complete paths

List every path satisfying the condition. Add their probabilities. This is the new denominator.
Keep the paths satisfying target and condition. Add these probabilities for the numerator.
Divide. Conditional probability is target-and-condition paths over all condition paths.

Worked example

A coin chooses how many marbles are drawn

A biased coin has $P(H)=1/3$ . A bag contains 4 red and 5 blue marbles. After H, draw one; after T, draw two without replacement. The probability of at least one red is $\frac13\frac49+\frac23\left(1-\frac59\frac48\right)$ . (9709/52/M/J/24 Q2)

Answer

\frac4{27}+\frac{13}{27}=\frac{17}{27}

Worked example

Reverse a three-stage tree

A white counter is chosen with probability $3/5$ and sends us to a box with 4 red and 3 blue marbles. A yellow counter is chosen with probability $2/5$ and sends us to a box with 5 red and 3 blue marbles. Two marbles are then drawn without replacement. Let E mean one red and one blue. (9709/53/O/N/25 Q5)

Find the success chance from each box. The two orders have equal products, so $P(E\mid W)=2(4/7)(3/6)=4/7$ and $P(E\mid Y)=2(5/8)(3/7)=15/28$ .
Weight the two boxes. The white-box contribution is $(3/5)(4/7)=12/35$ . The yellow-box contribution is $(2/5)(15/28)=3/14$ .
Add every path in the condition. $P(E)=12/35+3/14=39/70$ .

For $P(W\mid E)$ , the numerator keeps only the two successful paths that began with white. The denominator keeps all four paths in E.

Answer

P(W\mid E)=\frac{12/35}{39/70}

P(W\mid E)=\frac8{13}

Examiner note

Label branches before multiplying. A correct final decimal with no visible path structure may hide whether the changing denominator was handled correctly.

Your turnIndependent check [5 marks]

A bag contains 4 green and 3 yellow counters. Two are drawn without replacement. Find the probability of at least one green and the probability that the first counter was green, given that the two counters have different colours.

Show worked answer

$P(\text{at least one G})=1-\frac37\frac26=\frac67$ .

The different-colour paths are GY and YG. Their probabilities are $\frac47\frac36=\frac27$ and $\frac37\frac46=\frac27$ .

Let F mean “first is green” and D mean “different colours”.

Answer

P(\text{at least one G})=\frac67

P(F\mid D)=\frac12

Replacement decides whether the bag resets

With replacement: the first item returns. Counts and probabilities reset, so random repeated draws are independent.
Without replacement: one item has gone. The next denominator is one smaller and the relevant numerator may also change.
At every tree node, the outgoing branch probabilities must add to 1.

Worked example

Two reds without replacement

A bag contains 3 red and 2 blue counters.

Answer

P(RR)=\frac35\times\frac24=\frac3{10}

Common mistake

After drawing a red, the second red probability is

2/4

, not

3/5

. Both the total and the red count changed.

Multiply along a path; add alternative paths

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

A complete path joins events with AND, so multiply its branches.
Different successful paths are alternatives joined by OR, so add their path probabilities.
For “at least one”, the complement of no successes is often shorter.

Worked example

One red and one blue in either order

The successful paths are RB and BR.

P(RB)=\frac35\frac24=\frac3{10}

P(BR)=\frac25\frac34=\frac3{10}

Answer

\frac3{10}+\frac3{10}=\frac35

A tree condition is built from complete paths

List every path satisfying the condition. Add their probabilities. This is the new denominator.
Keep the paths satisfying target and condition. Add these probabilities for the numerator.
Divide. Conditional probability is target-and-condition paths over all condition paths.

Worked example

A coin chooses how many marbles are drawn

Answer

\frac4{27}+\frac{13}{27}=\frac{17}{27}

Worked example

Reverse a three-stage tree

Find the success chance from each box. The two orders have equal products, so $P(E\mid W)=2(4/7)(3/6)=4/7$ and $P(E\mid Y)=2(5/8)(3/7)=15/28$ .
Weight the two boxes. The white-box contribution is $(3/5)(4/7)=12/35$ . The yellow-box contribution is $(2/5)(15/28)=3/14$ .
Add every path in the condition. $P(E)=12/35+3/14=39/70$ .

For $P(W\mid E)$ , the numerator keeps only the two successful paths that began with white. The denominator keeps all four paths in E.

Answer

P(W\mid E)=\frac{12/35}{39/70}

P(W\mid E)=\frac8{13}

Examiner note

Label branches before multiplying. A correct final decimal with no visible path structure may hide whether the changing denominator was handled correctly.

Your turnIndependent check [5 marks]

Show worked answer

$P(\text{at least one G})=1-\frac37\frac26=\frac67$ .

The different-colour paths are GY and YG. Their probabilities are $\frac47\frac36=\frac27$ and $\frac37\frac46=\frac27$ .

Let F mean “first is green” and D mean “different colours”.

Answer

P(\text{at least one G})=\frac67

P(F\mid D)=\frac12