5.3 Probability
- Syllabus
- 9709–2028–2029
- Topic
- 5.3
- Level
- A2
Learning objectives
5.3.1Permutations and combinations• evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations e.g. the total score when two fair dice are thrown. e.g. drawing balls at random from a bag containing balls of different colours.5.3.2Probability rules• use addition and multiplication of probabilities, as appropriate, in simple cases Explicit use of the general formula AB AB ABPP PP,+ -= +^^ ^^hh hh is not required.5.3.3Exclusive and independent events• understand the meaning of exclusive and independent events, including determination of whether events A and B are independent by comparing the values of ABP +^h and ABPP #^^ hh5.3.4Elementary probability• calculate and use conditional probabilities in simple cases. e.g. situations that can be represented by a sample space of equiprobable elementary events, or a tree diagram. The use of AB B ABP P P +=^ ^ ^h h h may be required in simple cases.
Count the same kind of equally likely outcome
Forequiprobableelementaryoutcomes,P(A)=\frac{\text{number of outcomes in }A}{\text{total number of outcomes}}.Thetwocountsmustusethesameoutcomeunit.
| Situation | Build the probability |
|---|---|
| small sample space | enumerate every equally likely elementary outcome |
| ordered choices | count with permutations |
| unordered selections | count with combinations |
For two fair dice, use 36 ordered pairs. A total of 8 occurs for (2,6),(3,5),(4,4),(5,3),(6,2), so the probability is 5/36.
From 5 red and 3 blue balls, two chosen together have (28) equiprobable unordered pairs. Exactly one of each has (15)(13) pairs, giving 15/28.
Do not count favourable outcomes as ordered arrangements and total outcomes as unordered selections. First write what one elementary outcome means.
Multiply along a path and add alternative paths
| Event structure | Operation |
|---|---|
| successive stages on one path | multiply branch probabilities |
| mutually exclusive alternative paths | add path probabilities |
| at least one / not / neither | consider 1−P(complement) |
A bag has 3 red and 2 blue balls. Two are drawn without replacement. The path red then blue has probability 3/5×2/4=3/10; blue then red has probability 2/5×3/4=3/10. Therefore one of each has probability 3/10+3/10=3/5.
Without replacement, update both the favourable count and total after the first draw. With replacement, the second-stage probabilities return to their original values.
Do not add probabilities along one path or multiply alternative paths. The syllabus does not require explicit use of the general overlapping-events addition formula here.
Mutual exclusivity and independence describe different relationships
Events are mutually exclusive when A∩B=∅. They are independent when P(A∩B)=P(A)P(B), equivalently P(A|B)=P(A) when defined.
Disjoint non-zero events cannot be independent: learning that one occurred makes the other impossible. Test the stated relationship with the correct equation.
A single die roll being even and odd is mutually exclusive; two independent coin tosses are not mutually exclusive across different tosses.
“Independent” does not mean unrelated in everyday language, and mutually exclusive events are not independent unless one has probability zero.
Conditioning restricts the possible outcomes
If $P(B)>0$, thenP(A\mid B)=\frac{P(A\cap B)}{P(B)}.Read $A\mid B$ as ‘$A$ given that $B$ has occurred’.
| Representation | What the condition changes |
|---|---|
| equiprobable sample space | keep only outcomes satisfying the given event; this becomes the new denominator |
| tree diagram | start from the branch or branches compatible with the given information |
A fair die is known to show more than 2. The restricted outcomes are {3,4,5,6}. Given this information, the probability of an even result is 2/4=1/2, not 3/6.
A bag contains 5 red and 3 blue balls. If the first ball drawn without replacement is known to be red, then P(second red∣first red)=4/7. Equivalently, a joint probability can be built as P(A∩B)=P(B)P(A∣B).
The condition belongs after the vertical bar and determines the denominator. P(A∣B) and P(B∣A) are usually different.