01 · Red then blue
Find P(RB).
Hint
After red, all 3 blue counters remain among 6 counters.
Worked solution
(4/7)(3/6)=12/42=2/7.
Understand · explore · practise
Sampling without replacement
Update counts after each draw, calculate dependent tree paths, compare with replacement and distinguish unchanged marginal probabilities from independence.
Before you startTree diagrams, path multiplication and event addition.
01 / Update the bag after each result
The next probability depends on the path taken.
Without replacement, both the total count and the relevant colour count may change.
Write the remaining composition beside each branch point.
After red, the bag has 3 red and 3 blue. After blue, it has 4 red and 2 blue. Both bags contain 6 counters, but they give different second-stage probabilities.
Follow the changing bagExplore
Start with 4 red and 3 blue counters. Draw uniformly twice without replacement. R=red, B=blue. The second denominator is 6.
RR: (4/7)(3/6)=12/42=2/7. After a red draw only 3 red counters remain.
02 / Fill each node separately
Every node still has outgoing probabilities summing to 1.
Watch: the first result changes the second chance
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Start with 4 red and 3 blue.Worked example
First branches: R 4/7, B 3/7
There are 7 counters initially.
After R: R 3/6, B 3/6
A red counter has been removed.
After B: R 4/6, B 2/6
A blue counter has been removed instead.
02 · Two red
Find P(RR).
Hint
The red numerator also decreases.
Worked solution
(4/7)(3/6)=12/42=2/7. Using4/6 on the second branch would incorrectly retain the removed red.
03 / Keep the complete outcomes distinct
Different colours still has two orders.
03 · Blue then red
Find P(BR).
Hint
After blue there are 4 red out of 6.
Worked solution
(3/7)(4/6)=12/42=2/7.
04 · Two blue
Find P(BB).
Hint
Only 2 blue remain after the first blue.
Worked solution
(3/7)(2/6)=6/42=1/7.
05 · Different colours
Find P(different colours).
Hint
Add RB and BR.
Worked solution
2/7+2/7=4/7.
06 · Same colour
Find P(same colour).
Hint
Add RR and BB.
Worked solution
2/7+1/7=3/7. Same and different sum to 1.
04 / Use a short complementary event
At least one red excludes BB only.
07 · At least one red
Find P(at least one red).
Hint
Complement two blues.
Worked solution
1−1/7=6/7.
08 · Second draw red
Find P(second draw red) from the complete paths.
Hint
RR and BR both end red.
Worked solution
2/7+2/7=4/7. The marginal equals the original red proportion, even though the draws are dependent.
05 / Equal marginals do not imply independence
Conditioning on the first result changes the second chance.
P(second red)=4/7, but P(second red given first red)=3/6.
Since these differ, the colour events are dependent.
09 · Product test
Use the product definition to verify dependence of first-red and second-red.
Hint
Compare P(RR) with(4/7)(4/7).
Worked solution
P(RR)=2/7=14/49, whereas the product of marginals is16/49. They are unequal.
10 · Replacement comparison
Compare the different-colours probabilities with and without replacement.
Hint
Use 24/49 with replacement and 4/7 without.
Worked solution
Without replacement gives 28/49 rather than 24/49, larger by 4/49. Removing the first colour changes the balance towards the other colour.
06 / An alternative exact sample space
Label the physical counters to check the tree.
Imagine the seven counters have distinct identity labels.Worked example
There are 7×6=42 ordered distinct-counter pairs
The same physical counter cannot be selected twice.
RR has 4×3=12 pairs
Choose the first red identity and then a different red identity.
RB has 4×3=12; BR has 3×4=12; BB has 3×2=6
Their counts add to 42 and reproduce the tree probabilities.
11 · Counting error
Why is7×7 not the right total here?
Hint
It permits selecting the same physical counter twice.
Worked solution
Without replacement that counter is unavailable on the second draw. There are 6 second identities for each first identity, so 42 total.
07 / Find an unknown bag composition
A valid solution must be a whole-number count.
A different bag has r red and 2 blue. Two draws without replacement have P(RR)=2/5.Worked example
r/(r+2) × (r−1)/(r+1)=2/5
The second total and red count each decrease by 1.
5r(r−1)=2(r+2)(r+1)
The denominators are positive for r≥2.
3r²−11r−4=(3r+1)(r−4)=0
Only r= 4 is a valid count; reject −1/3.
12 · Check the bag
Verify r= 4 in the original probability expression.
Hint
The new bag has 6 counters altogether.
Worked solution
(4/6)(3/5)=12/30=2/5.
13 · One red only
What is P(RR) if the original bag instead has 1 red and 6 blue?
Hint
The second red branch after red has probability 0.
Worked solution
(1/7)(0/6)=0. Two reds are impossible without replacement.
08 / Three colours and three draws
A longer history changes the next composition.
A different bag has3 red,2 blue and1 green. Draw three without replacement.Worked example
For RBG: (3/6)(2/5)(1/4)=1/20
After red then blue,2 red,1 blue and1 green remain.
There are six orders with one of each colour
RBG, RGB, BRG, BGR, GRB and GBR are disjoint.
Each order has probability6/(6×5×4)=1/20
Their total is6/20=3/10. Equality here follows from the product of the available counts, not from assuming all colour sequences are equally likely.
14 · Next colour after a history
After R then B in this three-colour bag, find P(next red) and P(next green).
Hint
Write the remaining composition.
Worked solution
There are2 red,1 blue and1 green, total4. Probabilities are2/4=1/2 and1/4 respectively.
15 · Exactly two red and one blue
Find the probability of exactly two reds and one blue in the three draws.
Hint
List RRB,RBR,BRR.
Worked solution
Each path has probability(3×2×2)/(6×5×4)=1/10. The three disjoint orders give3/10.
16 · Colour sequences not equal
Compare RRR with RBG. Are these two colour sequences equally likely?
Hint
The numbers of physical-counter realisations differ.
Worked solution
P(RRR)=(3×2×1)/(6×5×4)=1/20, equal to RBG here, but that does not make all colour sequences equal: RRB has1/10 and GGG is impossible. Enumerate or multiply each path from the actual counts.
09 / Track what remains
The branch labels describe the bag after the previous outcomes.
Update the counts at every node. Multiply the conditional branch probabilities along a path and add disjoint complete paths for an event. Use labelled-counter enumeration as a check and do not infer independence merely from matching marginal probabilities.
Section 1 of 9 · Update the bag after each result