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List outcomes systematically, distinguish events from outcomes and calculate probabilities only when the equal-likelihood assumptions justify counting.
Before you startFractions, simple sets and basic probability vocabulary.
01 / Describe the random experiment
Sample space: all possible outcomes. Event: the outcomes satisfying a condition.
For equally likely outcomes, P(event) = number in event ÷ total number of outcomes.
First decide what is recorded and what counts as one outcome. A probability lies between 0 and 1. In a finite equally likely sample space, the empty event has probability 0 and the whole sample space has probability 1.
Spinner A is fair with labels 1, 2, 3. Spinner B is fair with labels 1, 2, 3, 4. Assume independent spins. Each ordered pair (A, B) has probability 1/12.
Sum 5 occurs for (1,4), (2,3), (3,2): 3 of 12 equally likely outcomes, so probability = 1/4.
02 / Keep ordered outcomes separate
There are 3 × 4 = 12 ordered pairs
List each A value with each B value exactly once.
(1,3) and (3,1) are different outcomes
The labels belong to different spinners, even though their sums match.
Fairness plus independence makes each pair equally likely
Each has probability (1/3)×(1/4) = 1/12.
A fair five-sided spinner and an independent fair two-sided spinner are used. How many equally likely ordered pairs are there?
Each of the five first results has two possible second results.
5×2 = 10 ordered pairs, each with probability 1/10.
In the 3-by-4 experiment, are (2,3) and (3,2) the same outcome?
Which spinner produced which label?
No. They have the same sum but different ordered results. Both must be included if the event requires sum 5.
03 / Count the outcomes that qualify
List the pairs giving sum 5 and find the probability.
Check each possible A label.
The pairs are (1,4), (2,3), (3,2). The probability is 3/12 = 1/4.
Find the probability that the product is even.
The product is odd only when both labels are odd.
There are 2 odd A labels and 2 odd B labels, giving 4 odd-product pairs. The other 8 pairs have even product, so the probability is 8/12 = 2/3.
04 / Do not count different sums as equal chances
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Their outcome counts are 1, 2, 3, 3, 2 and 1
Each count comes from the equally likely ordered pairs.
P(sum = 2) = 1/12; P(sum = 5) = 3/12
Different possible sums do not have equal probabilities.
Six possible sum values does not imply a probability of 1/6 each
Count the underlying equally likely outcomes or use their known probabilities.
A learner says P(sum = 5) = 1/6 because six sums are possible. Explain the mistake.
Are the six sums equally likely?
No. Sum 5 has three underlying pairs while sum 2 has only one. The equally likely objects are the twelve pairs, giving P(sum = 5) = 1/4.
Find P(sum = 7).
Only the largest labels can total 7.
Only (3,4) qualifies, so P(sum = 7) = 1/12.
05 / Use everything outside the event
P(not E) = 1 − P(E).
Every outcome is either in E or outside E, with no overlap.
What is the probability that the spinner labels do not match?
Matching pairs are (1,1), (2,2), (3,3).
P(match) = 3/12 = 1/4, so P(not match) = 3/4.
In this experiment, find P(sum = 9) and P(sum ≤ 7).
Check the largest possible sum.
The largest sum is 7. Therefore P(sum = 9) = 0 and P(sum ≤ 7) = 1.
06 / Check why counting is valid
Two coins always show the same face: HH or TT, each with probability 1/2. Each coin is individually fair. Are the four ordered coin outcomes equally likely?
Inspect which outcomes can actually occur.
No. HT and TH have probability 0. The coins are dependent. Individual fairness alone does not make the joint outcomes equally likely.
Now A has probabilities 1/2, 1/3, 1/6 for labels 1,2,3. B remains fair and independent. Find P(sum > 5).
Add probabilities of (2,4), (3,3), (3,4), rather than assigning each 1/12.
P = (1/3)(1/4) + (1/6)(1/4) + (1/6)(1/4) = 1/12 + 1/24 + 1/24 = 1/6. The old equal-count answer 1/4 is not valid for this changed model.
07 / Check your sample space
Find P(A < B) for the original fair independent spinners.
For A = 1,2,3, count larger B values.
There are 3 + 2 + 1 = 6 qualifying pairs, so P(A < B) = 1/2. Equality does not satisfy a strict inequality.
Find the probability that the labels differ by exactly 1.
Use absolute difference; either label can be larger.
The five pairs are (1,2), (2,1), (2,3), (3,2), (3,4). Probability = 5/12.
A calculation gives probability 14/12 in this twelve-outcome model. What must be checked?
An event cannot contain more distinct outcomes than the sample space.
The result is invalid. Check duplicate counting, overlaps and the denominator; a probability must lie between 0 and 1.
Does the exact model probability 1/4 mean exactly one quarter of outcomes in every short run will satisfy sum 5?
A probability describes chance, not a quota.
No. Short-run counts vary. The value 1/4 is exact for the stated probability model, not a guaranteed proportion in each finite sequence.
08 / Model before counting
Define the experiment, list each outcome once, identify the event and check equal likelihood. Distinguish ordered pairs from derived values such as sums. Use complements when simpler and state the independence or fairness assumptions your calculation needs.
Section 1 of 8 · Describe the random experiment