01 · Both red
Find P(RR).
Hint
Use the red branch twice.
Worked solution
(4/7)(4/7)=16/49.
Understand · explore · practise
Tree diagrams with replacement
Read successive probability branches, multiply along complete paths, add distinct paths for an event and use complements with independent repeated sampling.
Before you startFractions, complementary events and independence.
01 / One branch per possible next outcome
A tree records the order of events.
Branches leaving any one node sum to 1.
A complete path describes one ordered outcome across all stages.
Replacement restores the bag composition. Independent uniform draws also require the stated mixing and selection model. In this experiment, both stages have red probability 4/7 and blue probability 3/7.
Select complete pathsExplore
A bag contains 4 red and 3 blue counters. Draw uniformly, replace and mix, then draw again independently. R=red, B=blue.
RR: (4/7)(4/7)=16/49. Multiply along this one path.
02 / Multiply along a path
The second branch is a probability after the first result.
Watch: one red can happen in two orders
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Find the probability of red then blue.Worked example
First red: 4/7
Choose the first red branch.
Then blue: 3/7
Replacement restores the same blue proportion.
P(RB)=(4/7)(3/7)=12/49
This is one specific order, not the full different-colours event.
02 · Both blue
Find P(BB).
Hint
Use the blue branch twice.
Worked solution
(3/7)(3/7)=9/49.
03 / Add alternative complete paths
Exactly one red has two different orders.
P(different colours)=P(RB)+P(BR).
Complete ordered paths are mutually exclusive, so their probabilities add.
03 · Different colours
Find the probability of different colours.
Hint
Both RB and BR qualify.
Worked solution
12/49+12/49=24/49. Counting only RB misses the other order.
04 · Same colour
Find the probability of the same colour.
Hint
RR or BB, not both on this trial.
Worked solution
16/49+9/49=25/49.
05 · Whole tree
Verify the four path probabilities sum to 1.
Hint
Include RR,RB,BR,BB.
Worked solution
(16+12+12+9)/49=49/49=1.
04 / At least one versus exactly one
Choose the simpler event to calculate.
At least one red means RR, RB or BR.Worked example
The only excluded path is BB
Use the complement of no reds.
P(at least one red)=1−9/49=40/49
RR must be included.
Exactly one red remains 24/49
The two phrases describe different events.
06 · At least one blue
Find P(at least one blue).
Hint
Exclude only RR.
Worked solution
1−16/49=33/49.
07 · Second result
Find the probability the second draw is red by adding paths.
Hint
Both RR and BR end in red.
Worked solution
16/49+12/49=28/49=4/7, agreeing with the marginal probability.
05 / Independent stages can have different probabilities
Independence does not require identical chances.
An independent two-stage model has success chance 0.6 first, 0.3 second.Worked example
Both succeed: 0.6×0.3=0.18
Use the appropriate probability for each stage.
Exactly one succeeds: 0.6×0.7+0.4×0.3=0.54
The two alternative path probabilities need not match.
Neither succeeds: 0.4×0.7=0.28
The four paths 0.18,0.42,0.12,0.28 sum to 1.
08 · At least one success
Find P(at least one success) for this model.
Hint
Complement neither.
Worked solution
1−0.28=0.72.
09 · Equal mixed paths?
Why is it wrong here to double 0.6×0.7 for exactly one success?
Hint
Compare success-failure with failure-success.
Worked solution
The other order has probability 0.4×0.3=0.12, not0.42. The correct sum is0.42+0.12=0.54.
06 / Use a tree to find a missing chance
Keep probability constraints when solving.
10 · Find red probability
Two independent identical draws have P(RR)=0.36. Find the red probability p per draw.
Hint
Solve p²=0.36 with 0≤p≤1.
Worked solution
p=0.6. Reject −0.6 because a probability cannot be negative.
11 · None given
Two independent identical trials have P(no successes)=0.49. Find the success probability p.
Hint
(1−p)²=0.49 and 1−p≥0.
Worked solution
1−p=0.7, so p=0.3.
07 / Keep the branching model consistent
Each node needs its own complete alternatives.
12 · Branch-sum mistake
A node has only two exhaustive branches labelled 0.4 and 0.5. What is wrong?
Hint
Outgoing branches must sum to 1.
Worked solution
They sum to 0.9. Either a branch is missing or a probability is wrong.
13 · Replacement matters
Would the second red probability still be 4/7 after drawing red without replacement?
Hint
Update the physical bag contents.
Worked solution
No. There would be 3 red out of 6 counters, giving 1/2. The present tree applies to replacement and independent mixing.
14 · Add or multiply?
Explain why P(RB) multiplies but P(RB or BR) adds.
Hint
Distinguish stages within an outcome from alternative outcomes.
Worked solution
RB requires successive results along one path, so multiply the branch probabilities. RB and BR are disjoint complete outcomes that both satisfy the event, so add their path probabilities.
08 / Read paths before calculating
Branch probabilities, path probabilities and event probabilities play different roles.
Check that branches leaving each node sum to 1. Multiply along a complete path using the correct probability after each stage. Add the disjoint paths that satisfy the event. Use complements for at least one when that reduces the work.
Section 1 of 8 · One branch per possible next outcome