Sample space and events
Probability: How likely something is: a number from 0 (impossible) to 1 (certain).
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Trial and outcome: Picking one student is a trial. "Walks" is one possible outcome.
Check two things before you divide: 1. U holds every possible outcome: all 25 students, none left out.
2. Every outcome is equally likely: each student has the same chance of being picked.
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Calculating probabilities
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Exam-style question [2 marks]
A month of the year is chosen at random.
(a) Find the probability that its name starts with M.
(b) Find the probability that its name does not start with M.
Solution
- List U: the 12 months, each equally likely.
- A = {March, May}.
- (a) Divide.
- (b) Not A: subtract from 1.
Final answer
(a) 1/6 ≈ 0.167 (b) 5/6 ≈ 0.833
Exam tip: Give the answer as a fraction (2/12 or 1/6), a decimal (0.167, to 3 significant figures) or a percentage (16.7%): all get full marks.
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Complementary events
Complementary: Events covering all: P(A) + P(A′) = 1.
Example: a flight is late, or it is not late.
Worked example
P(rain)=0.3.
Find P(no rain).
Solution
- Rain and no rain cover every day, and cannot both happen.
- P(no rain) = 1 − 0.3 = 0.7
Final answer
0.7
"Not A" in words: Exam questions often ask which wording means A′. A′ is everything that is not in A.
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Two dice: the 36-outcome grid
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Two dice: counting pairs in the exam
No grid? Go through the first die: For one event, try each value of the first die and ask what the second die must be. It must be 1 to 6. This way no pair is missed, including the reversed one.
| First die | Second die needed (11 − first) | Possible? |
|---|---|---|
| 1 | 10 | no |
| 2 | 9 | no |
| 3 | 8 | no |
| 4 | 7 | no |
| 5 | 6 | yes: (5, 6) |
| 6 | 5 | yes: (6, 5) |
Another check: product = 12: Same method, different rule: the second die must be 12 ÷ first die, and it must be a whole number from 1 to 6.
| First die | Second die needed (12 ÷ first) | Possible? |
|---|---|---|
| 1 | 12 | no |
| 2 | 6 | yes: (2, 6) |
| 3 | 4 | yes: (3, 4) |
| 4 | 3 | yes: (4, 3) |
| 5 | 2.4 | no (not a whole number) |
| 6 | 2 | yes: (6, 2) |
P(product = 12): 4 pairs work: (2, 6), (3, 4), (4, 3), (6, 2). So P(product = 12) = 4/36 = 1/9.
One event, or several?: Only (5, 6) and (6, 5) work, so P(sum = 11) = 2/36.
One event: go through the first die like this. Several events in one question (e.g. P(sum = 7), P(sum = 8) and P(sum ≥ 10)): draw the 6 × 6 grid once and count each from it.
Exam-style question [2 marks]
Two fair six-sided dice are rolled.
(a) Find the probability that the sum is at least 10.
(b) Find the probability that the product of the two numbers is 6.
Solution
- U: 36 equally likely pairs.
- (a) Count in the sum grid: 10, 11 or 12 sits in 6 cells, (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6).
- (b) Go through the first die: the second must be 6 ÷ first, a whole number from 1 to 6. 1 → 6 yes, 2 → 3 yes, 3 → 2 yes, 4 → 1.5 no, 5 → 1.2 no, 6 → 1 yes. So (1, 6), (2, 3), (3, 2), (6, 1): 4 pairs.
Final answer
(a) 1/6 ≈ 0.167 (b) 1/9 ≈ 0.111
Exam tip: Count the pairs, not the totals. A sum of 7 can be made 6 ways: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). A sum of 12 can be made only 1 way: (6, 6). So P(sum = 7) = 6/36 and P(sum = 12) = 1/36.