Probability errors often begin before any fraction is calculated. The real decision is structural: are events happening as alternatives, in sequence, with or without replacement, independently, or as a complement of something easier to count?
This fiftieth Secondary 4 Mathematics Learning Guide develops probability as event logic rather than a collection of isolated rules. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
It deepens Probability, Combined Events and Tree Diagrams by focusing on the decision layer: what operation is justified by the event structure.
The first question: OR, AND, NOT, or THEN?
- OR often suggests addition, with overlap handled if necessary.
- AND or sequential outcomes often suggest multiplication along a path.
- NOT, “none”, or “at least one” may be easier through a complement.
- THEN may change the sample space if the first event alters what remains.
Choose the operation only after identifying the event relationship.
Mutually exclusive events cannot happen together
If events A and B are mutually exclusive, P(A and B)=0. Therefore:
P(A or B)=P(A)+P(B).
Worked Example 1 | Add mutually exclusive outcomes
A fair six-sided die is rolled. Find the probability of rolling a 1 or a 5.
P(1 or 5)=1/6+1/6=1/3.
The events cannot happen on the same single roll.
If events overlap, simple addition double-counts the overlap
For events that can occur together:
P(A or B)=P(A)+P(B)−P(A and B).
Worked Example 2 | Correct for overlap
From integers 1 to 20, one number is chosen at random. Find the probability it is a multiple of 2 or a multiple of 3.
Multiples of 2: 10. Multiples of 3: 6. Multiples of both 2 and 3 are multiples of 6: 3.
P= (10+6−3)/20=13/20.
Multiplication follows a path
When an outcome requires several stages in sequence, multiply the probabilities along that path.
Worked Example 3 | Independent sequence
A fair coin is tossed twice. Find P(head then head).
1/2×1/2=1/4.
The first toss does not change the probability on the second toss, so the events are independent.
Independent does not mean mutually exclusive
Independent events can happen together without one changing the probability of the other. Mutually exclusive events cannot happen together at all.
For events with positive probability, mutually exclusive events are not independent because if one occurs, the other becomes impossible.
Worked Example 4 | Distinguish the ideas
A die is rolled. Let A be “even” and B be “greater than 4”.
A={2,4,6}; B={5,6}. They overlap at 6, so they are not mutually exclusive.
P(A)=1/2, P(B)=1/3, P(A and B)=1/6.
P(A)P(B)=1/2×1/3=1/6.
Since P(A and B)=P(A)P(B), A and B are independent in this example.
Without replacement changes the denominator
If an object is removed and not returned, the number of available outcomes changes. The second probability must be recalculated from the new sample space.
Worked Example 5 | Change the denominator
A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement. Find P(two red).
P(RR)=5/8×4/7=5/14.
The denominator changes from 8 to 7 because only seven counters remain after the first draw.
With replacement restores the original sample space
If the first counter is returned before the second draw, the bag again contains all eight counters.
Worked Example 6 | With replacement
Using the same bag, find P(two red) with replacement.
5/8×5/8=25/64.
With replacement and without replacement are different mathematical models of the same physical bag.
Complements are often the shortest route
Since total probability is 1:
P(not A)=1−P(A).
This is especially useful for “at least one” when counting many success cases would be long.
Worked Example 7 | At least one success
A fair coin is tossed three times. Find the probability of at least one head.
Complement is no heads, meaning TTT.
P(at least one H)=1−(1/2)³=7/8.
“Exactly one” is not the same as “at least one”
Exactly one success requires counting only paths with one success. At least one includes one, two, three or more successes depending on the number of trials.
Worked Example 8 | Exactly one head
A fair coin is tossed twice. Find P(exactly one head).
Valid paths: HT and TH.
P=1/4+1/4=1/2.
Tree diagrams make path logic visible
Multiply along branches because one complete path requires all events on that path. Add across separate successful paths because any one of those paths satisfies the target event.
Multiply along. Add across.
Worked Example 9 | Two successful paths
A bag contains 2 red and 3 blue counters. Two are drawn without replacement. Find P(one red and one blue).
There are two paths: RB and BR.
- P(RB)=2/5×3/4=3/10.
- P(BR)=3/5×2/4=3/10.
Total=3/5.
Unknown probabilities can be solved algebraically
If an event probability is represented by x, use the same event logic to build an equation, then filter any algebraic roots through the probability constraint 0≤x≤1.
Worked Example 10 | Solve an unknown event probability
An event A has probability x. Its complement has probability 0.35. Find x.
x+0.35=1, so x=0.65.
Experimental probability is evidence, not certainty
If an event occurs 47 times in 100 trials, the experimental probability is 0.47. This can estimate a long-run probability, but it does not prove the theoretical probability is exactly 0.47.
Worked Example 11 | Use experimental frequency
A component fails 8 times in 400 tests. Estimate the probability of failure.
8/400=0.02.
Estimated failure rate=2% under the observed testing conditions.
Probability decision table
| Wording / structure | Likely move |
|---|---|
| A or B, mutually exclusive | Add |
| A or B, overlap possible | Add then subtract overlap |
| A then B | Multiply along the path |
| Without replacement | Update numerator and denominator |
| With replacement | Restore original sample space |
| At least one | Consider 1−P(none) |
| Exactly one | Add all one-success paths only |
| Not A | 1−P(A) |
Common failure modes
| Failure | Cause | Repair |
|---|---|---|
| Adds probabilities for sequential events | OR and AND confused | Map the event path first |
| Multiplies mutually exclusive alternatives | Alternative paths confused with sequence | Add separate successful outcomes |
| Keeps denominator fixed without replacement | Sample space not updated | Recount what remains |
| Uses complement for exactly one | At least one and exactly one confused | List target paths |
| Calls mutually exclusive events independent | Definitions blended | Ask whether both can occur and whether one changes the other |
| Forgets overlap when adding | Double-counting | Subtract intersection once |
Independent practice
- A fair die is rolled. Find P(2 or 6).
- One number from 1 to 12 is chosen. Find P(multiple of 2 or multiple of 3).
- A fair coin is tossed three times. Find P(all tails).
- A bag has 4 red and 5 blue counters. Find P(two red without replacement).
- Using the same bag, find P(two red with replacement).
- A fair die is rolled twice. Find P(at least one 6).
- A bag has 3 green and 2 yellow counters. Two are drawn without replacement. Find P(exactly one green).
Explained answers
1. 1/6+1/6=1/3.
2. Multiples of 2: 6; multiples of 3: 4; overlap multiples of 6: 2. Probability=(6+4−2)/12=2/3.
3. (1/2)³=1/8.
4. 4/9×3/8=1/6.
5. 4/9×4/9=16/81.
6. 1−(5/6)²=11/36.
7. GY or YG: 3/5×2/4 + 2/5×3/4=3/5.
Final thought
Probability becomes far more reliable when operations follow event structure. Addition, multiplication, complements and changing denominators are consequences of how the sample space behaves, not buttons to press because a keyword appeared.
Draw the event logic first. Then let the structure tell you whether to add, multiply, subtract from one or change what remains.
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