Secondary Mathematics: Probability Trees, Conditional Events and Without-Replacement Reasoning

Secondary Mathematics · Worked Repair Guide 29

Multi-stage probability becomes reliable when each branch represents a condition that is actually true at that stage. A tree diagram is not decoration. It is a record of how the sample space changes after an outcome has occurred.

This guide develops one central habit: update the available outcomes before writing the next probability. If an object is replaced, later probabilities may stay the same. If it is not replaced, the denominator and sometimes the favourable count change. Conditional probability is therefore a question about the reduced situation after information becomes known.

All bags, cards, spinners and selection examples below are invented teaching material. Use Probability, Sample Spaces and Independence for the foundational route and Why Does Probability Feel So Counterintuitive? for the shorter diagnostic explanation.

1. Probabilities belong to a defined sample space

A fair six-sided die has sample space {1,2,3,4,5,6}. The probability of an even result is 3/6=1/2.

If we are told the result is greater than 3, the relevant conditional sample space becomes {4,5,6}. Within that reduced space, the probability the result is even is 2/3.

Entry check: P(even)=1/2, but P(even | greater than 3)=2/3. New information can change probability even though the die itself has not changed.

2. Complements are often the shortest route

For an event A, P(not A)=1−P(A).

If a single trial succeeds with probability 0.3, failure has probability 0.7.

For repeated independent trials, “at least one success” is often easiest through its complement “no successes”.

For two independent trials with success probability 0.3, P(at least one success)=1−0.7²=0.51.

3. Tree diagrams multiply along a route

Suppose an invented machine labels an item red with probability 0.4 and blue with probability 0.6 on each independent run.

The probability of red then blue is 0.4×0.6=0.24.

The multiplication represents successive conditions: take the red branch, then from that outcome take the blue branch.

Do not add probabilities along a single route. Addition is used later when combining separate mutually exclusive routes to the same requested event.

4. Add separate routes that lead to the same final event

For two independent red/blue runs with P(R)=0.4 and P(B)=0.6, exactly one red can occur by two routes: RB or BR.

P(RB)=0.4×0.6=0.24 and P(BR)=0.6×0.4=0.24.

Therefore P(exactly one red)=0.24+0.24=0.48.

The routes are mutually exclusive because a two-run sequence cannot be both RB and BR at the same time.

5. Replacement can preserve branch probabilities

A bag contains 3 red and 2 blue counters. One counter is selected, replaced, then another is selected.

Because the first counter is returned, each draw begins from 3 red and 2 blue. Thus P(R)=3/5 and P(B)=2/5 on both stages.

P(two red)=3/5×3/5=9/25.

Under this model, the second draw is independent of the first because the composition has been restored.

6. Without replacement changes the second stage

Now use the same bag but do not replace the first counter.

If the first counter is red, 2 red and 2 blue remain out of 4 total. So P(R second | R first)=2/4=1/2.

If the first counter is blue, 3 red and 1 blue remain. So P(R second | B first)=3/4.

The second-stage probabilities depend on the first outcome. The events are therefore not independent.

7. Work a complete without-replacement tree

Bag: 3 red, 2 blue. Draw two without replacement.

P(RR)=3/5×2/4=6/20=3/10.

P(RB)=3/5×2/4=3/10.

P(BR)=2/5×3/4=6/20=3/10.

P(BB)=2/5×1/4=2/20=1/10.

The four route probabilities sum to 1, providing a strong consistency check.

8. “At least one” often invites a complement

In the same without-replacement bag, find P(at least one red).

The complement is two blue. P(BB)=1/10.

Therefore P(at least one red)=1−1/10=9/10.

Adding RR, RB and BR also gives 3/10+3/10+3/10=9/10, an independent check.

9. Conditional probability restricts the denominator

The notation P(A|B) means “probability of A given B”. When P(B)>0, P(A|B)=P(A∩B)/P(B).

Suppose 40 invented observations include 18 in group A, 12 in both A and B, and 20 in group B.

Given that an observation is in B, the relevant denominator is the 20 B observations. Twelve of those are also in A.

Therefore P(A|B)=12/20=3/5.

10. The order in conditional probability matters

Using the same counts, P(B|A)=12/18=2/3.

P(A|B)=3/5 and P(B|A)=2/3 are different because the conditioning group changes.

The vertical bar is directional: the event after the bar specifies the restricted sample space.

Do not assume conditional probabilities are symmetric.

11. Independence can be tested through conditional probability

Events A and B are independent when knowing B does not change the probability of A. One equivalent test, when P(B)>0, is P(A|B)=P(A).

Another equivalent test is P(A∩B)=P(A)P(B).

If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.2, then 0.4×0.5=0.2, so the given values are consistent with independence.

Independence is a relationship between events, not a visual property of a diagram.

12. Mutually exclusive is not the same as independent

Mutually exclusive events cannot occur together, so P(A∩B)=0.

If both have positive probability, they cannot be independent because independence would require P(A∩B)=P(A)P(B)>0.

For one die roll, “even” and “odd” are mutually exclusive. Knowing the result is odd changes P(even) from 1/2 to 0.

Do not use “independent” as another word for “separate”.

13. Three-stage trees require continuous updating

A bag has 2 red and 3 blue counters. Three counters are drawn without replacement.

P(RRB)=2/5×1/4×3/3=1/10.

P(RBR)=2/5×3/4×1/3=1/10.

P(BRR)=3/5×2/4×1/3=1/10.

Thus P(exactly two red)=3/10. The three route probabilities happen to be equal here, but that equality should be calculated rather than assumed.

14. Order can matter even when the final composition does not

When sampling without replacement, the probability of a particular composition can be obtained by adding all relevant orders.

In the 2-red,3-blue bag above, “exactly two red in three draws” includes RRB, RBR and BRR.

The final count of colours is the same, but the tree records different paths to reach it.

This is a useful bridge to combinatorial counting in more advanced probability work.

15. Expected frequency is not a guarantee

If an event has probability 0.3 and a model is repeated 200 times under comparable conditions, expected frequency is 200×0.3=60.

This does not promise exactly 60 occurrences. It is a long-run model expectation.

Observed frequency may be above or below 60.

This distinction connects to A Probability of 70% Does Not Promise Seven Successes Out of the Next Ten.

16. Reverse tree problems solve for an unknown branch probability

Suppose a first-stage event A has probability p. Given A, success S has probability 0.8; given not A, success has probability 0.3. Suppose overall P(S)=0.5.

Then 0.8p+0.3(1−p)=0.5.

0.8p+0.3−0.3p=0.5, so 0.5p=0.2 and p=0.4.

The two success routes are mutually exclusive and together account for every way S can occur under the stated model.

17. Probability statements require compatible denominators

When combining fractions from different conditional branches, do not compare numerators without their denominators.

For example, “2 red remain” means different probabilities depending on whether 4 or 5 total objects remain.

Always write favourable/possible for the current branch.

This is the probability version of the “correct base” principle from ratio and percentage work.

18. Capstone: combine complement, dependence and conditioning

An invented box contains 4 green and 3 yellow tokens. Two are drawn without replacement.

P(two yellow)=3/7×2/6=1/7.

Therefore P(at least one green)=1−1/7=6/7.

Given that the first token is yellow, P(second green)=4/6=2/3.

Unconditionally, P(second green)=4/7 by symmetry of positions in sampling without replacement. The conditional value 2/3 is different because knowing a yellow was removed changes the remaining composition.

19. Independent practice

1. A fair die is rolled. Find P(even).
2. Given the die result is greater than 3, find P(even).
3. An event has probability 0.27. Find its complement.
4. Two independent trials have success probability 0.4. Find P(two successes).
5. For the same trials, find P(exactly one success).
6. For the same trials, find P(at least one success).
7. A bag has 3 red and 2 blue counters. Two are drawn with replacement. Find P(RR).
8. Now draw without replacement. Find P(RR).
9. In Question 8, find P(BB).
10. In Question 8, find P(at least one red).
11. A bag has 2 red and 3 blue. Three are drawn without replacement. Find P(RRB).
12. For the same bag, find P(exactly two red).
13. If P(A)=0.5, P(B)=0.4 and P(A∩B)=0.2, test independence.
14. If A and B are mutually exclusive with P(A)=0.3 and P(B)=0.2, find P(A∪B).
15. Explain why the events in Question 14 are not independent.
16. Forty observations include 18 in A, 20 in B and 12 in both. Find P(A|B).
17. Using the same counts, find P(B|A).
18. An event with probability 0.35 is repeated 400 times. Find expected frequency.
19. Explain why expected frequency is not a guaranteed count.
20. A first-stage event has probability p. Success probability is 0.8 after the event and 0.3 otherwise. Overall success probability is 0.5. Find p.

20. Worked answers

1. 1/2.

2. 2/3. Conditional sample space {4,5,6}; even values 4,6.

3. 0.73.

4. 0.16. 0.4².

5. 0.48. 0.4×0.6+0.6×0.4.

6. 0.64. 1−0.6².

7. 9/25.

8. 3/10. 3/5×2/4.

9. 1/10. 2/5×1/4.

10. 9/10. Complement of BB.

11. 1/10. 2/5×1/4×3/3.

12. 3/10. Sum RRB, RBR and BRR.

13. Independent. 0.5×0.4=0.2.

14. 0.5. Mutually exclusive events add directly.

15. Because P(A∩B)=0 while P(A)P(B)=0.06, so the independence condition fails.

16. 3/5. 12/20.

17. 2/3. 12/18.

18. 140. 400×0.35.

19. Probability describes a distribution of possible outcomes across repetitions; an actual finite run can differ from the expected count.

20. p=0.4. Solve 0.8p+0.3(1−p)=0.5.

21. Diagnose the first probability update error

Common failures include keeping the original denominator after a counter has been removed, multiplying routes that should be added, adding probabilities along one route, reversing P(A|B) and P(B|A), or calling mutually exclusive events independent.

A useful correction note says “update the bag”, “multiply along, add across”, “event after the bar defines the denominator”, or “independence means information does not change probability”.

Then change whether replacement occurs. A secure learner should immediately predict whether the second-stage branch probabilities remain the same.

22. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Probability, Sample Spaces and Independence for foundations and Sets, Venn Diagrams and Logical Counting for event-set reasoning.

Within Batch 08, continue to Histograms, Frequency Density and Grouped Data, Distance–Time, Speed–Time and Motion Graphs, or Approximation, Significant Figures, Bounds and Error Intervals.

23. Sources and scope

The counters, probabilities, conditional tables and practice questions are original teaching material. Standard probability definitions are used in their ordinary mathematical sense.

For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match formal conditional-probability notation and multi-stage extensions to the learner’s actual subject level and school programme.