The shape of the lesson
By the end of the lesson, learners can
- Equally likely outcomes: In the lab's sample-space tab, roll two dice many times and compare the frequency of each total with the grid of 36 outcomes.
- Exclusive and independent: Two words that are easily confused.
- Tree diagrams: Use the lab's tree-diagram tab to build a 'without replacement' tree, then compare it with Fig. 1.
- Probability by combinations: When several items are picked at once, count the favourable selections and the total selections with \({}^{n}C_{r}\).
- Equally likely: favourable outcomes ÷ total outcomes.
- Or, exclusive: add. And, independent: multiply.
- Trees: multiply along a route, add the routes; check later branches for 'without replacement'.
- At least one: \(1 - \text{P}(\text{none})\).
- Several picked at once: \(\dfrac{\text{favourable selections}}{\text{total selections}}\) using \({}^{n}C_{r}\).
One picture every idea returns to
For equally likely outcomes, \(\text{P}(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of outcomes}}\). If \(A\) and \(B\) are exclusive, \(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B)\). Along a tree, multiply; across separate routes, add. \(\text{P}(\text{at least one}) = 1 - \text{P}(\text{none})\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Count it, add it, multiply it.. Paper 5 probability rests on a few moves: count equally likely outcomes, add for 'or' when events are exclusive, multiply along the branches of a tree for 'and', and use combinations when several items are picked at once. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Equally likely outcomes | In the lab's sample-space tab, roll two dice many times and compare the frequency of each total with the grid of 36 outcomes. Project tab 1 of the lab (Combined probability lab, sample space) and let a student drive it while the class predicts. Model the worked example Totals of two dice on the board, one line at a time. Your turn: Two fair dice are thrown. Find the probability that the product of the scores is odd. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Exclusive and independent | Two words that are easily confused. Model the worked example Using both rules on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Tree diagrams | Use the lab's tree-diagram tab to build a 'without replacement' tree, then compare it with Fig. 1. Project tab 2 of the lab (Combined probability lab, tree diagrams) and let a student drive it while the class predicts. Model the worked example Without replacement on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Probability by combinations | When several items are picked at once, count the favourable selections and the total selections with \({}^{n}C_{r}\). Model the worked example Exactly two red on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 22 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Probability by combinations, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Saying P(total 8) \(= \dfrac{1}{11}\) because there are 11 possible totals. | The totals 2 to 12 are NOT equally likely; only the 36 ordered pairs are. |
| Adding probabilities of events that are NOT exclusive. | With \(\text{P}(A) = 0.6\) and \(\text{P}(B) = 0.7\), 'adding' gives 1.3, which is impossible; the overlap has been counted twice. |
| Keeping the same denominators on the second branches. | Without replacement there are only 7 balls left, so the second-stage fractions are sevenths. |
| Forgetting the different orders when using a tree for 'exactly two red'. | RRB alone gives \(\dfrac{5}{28}\); there are 3 orders (RRB, RBR, BRR). Combinations count them automatically. |
Probability: sample spaces, exclusive events and tree diagrams: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Explain the difference between exclusive events and independent events, and state \(\text{P}(A \cap B)\) in each case.
Two fair dice are thrown. \(A\) is the event 'the total is 9' and \(B\) is the event 'the scores differ by 2'.
(a) Find \(\text{P}(A)\) and \(\text{P}(B)\). (b) Are \(A\) and \(B\) exclusive? (c) Find \(\text{P}(A \cup B)\).
A box contains 4 red and 6 green counters. Three counters are taken at random without replacement. Find the probability that (a) all three are the same colour, (b) at least one is red.
Ali and Bo take a driving test on the same day. The probability that Ali passes is 0.7 and the probability that Bo passes is 0.6, independently. Find the probability that (a) exactly one of them passes, (b) at least one of them passes.
Five people are chosen at random from 6 men and 4 women. Find the probability that more women than men are chosen.
Mia and Noor play a game. Mia throws a fair die first, then Noor, then Mia, then Noor. The first person to throw a six wins, and the game stops after four throws if nobody has thrown a six.
Find the probability that (a) Mia wins, (b) nobody wins, (c) Noor wins.
Answer key and marking notes
- Exclusive: \(A\) and \(B\) cannot both happen, so \(\text{P}(A \cap B) = 0\).
- Independent: the occurrence of one does not affect the probability of the other, so \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\).
- (a) Total 9: (3,6), (4,5), (5,4), (6,3), so \(\text{P}(A) = \tfrac{4}{36} = \tfrac{1}{9}\).
- Differ by 2: (1,3), (2,4), (3,5), (4,6) and their reverses, so \(\text{P}(B) = \tfrac{8}{36} = \tfrac{2}{9}\).
- (b) Two scores with total 9 differ by an odd number, so they cannot differ by 2. Yes, exclusive.
- (c) \(\tfrac{1}{9} + \tfrac{2}{9} = \tfrac{1}{3}\).
- (a) RRR: \(\tfrac{4}{10} \times \tfrac{3}{9} \times \tfrac{2}{8} = \tfrac{24}{720}\). GGG: \(\tfrac{6}{10} \times \tfrac{5}{9} \times \tfrac{4}{8} = \tfrac{120}{720}\).
- Same colour: \(\tfrac{144}{720} = \tfrac{1}{5}\).
- (b) \(1 - \text{P}(GGG) = 1 - \tfrac{120}{720} = \tfrac{5}{6}\).
- (a) \(0.7 \times 0.4 + 0.3 \times 0.6 = 0.28 + 0.18 = 0.46\).
- (b) \(1 - 0.3 \times 0.4 = 1 - 0.12 = 0.88\).
- More women than men means 3 women and 2 men, or 4 women and 1 man.
- \({}^{4}C_{3}\,{}^{6}C_{2} + {}^{4}C_{4}\,{}^{6}C_{1} = 4 \times 15 + 1 \times 6 = 66\).
- Total \({}^{10}C_{5} = 252\). Probability \(= \tfrac{66}{252} = \tfrac{11}{42} = 0.262\) (3 s.f.).
- (a) Mia wins on throw 1: \(\tfrac{1}{6}\). On throw 3: \(\left(\tfrac{5}{6}\right)^2 \tfrac{1}{6} = \tfrac{25}{216}\).
- \(\tfrac{36}{216} + \tfrac{25}{216} = \tfrac{61}{216} = 0.282\) (3 s.f.).
- (b) No six in four throws: \(\left(\tfrac{5}{6}\right)^4 = \tfrac{625}{1296} = 0.482\) (3 s.f.).
- (c) Noor on throw 2: \(\tfrac{5}{36}\); on throw 4: \(\left(\tfrac{5}{6}\right)^3 \tfrac{1}{6} = \tfrac{125}{1296}\). Total \(\tfrac{180 + 125}{1296} = \tfrac{305}{1296} = 0.235\) (3 s.f.).
- Check: \(\tfrac{366}{1296} + \tfrac{625}{1296} + \tfrac{305}{1296} = 1\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Two fair six-sided dice are thrown. Find the probability that the total score is at most 4.
- A\(\dfrac{1}{6}\)
- B\(\dfrac{3}{11}\)
- C\(\dfrac{1}{9}\)
- D\(\dfrac{5}{18}\)
Events \(A\) and \(B\) are independent, with \(\text{P}(A) = 0.4\) and \(\text{P}(B) = 0.35\). Find \(\text{P}(A \cup B)\).
- A0.61
- B0.75
- C0.14
- D0.39
The probability that a bulb fails in its first year is 0.15, independently of other bulbs. A room has 5 such bulbs. Find the probability that at least one of them fails in its first year.
- A0.556
- B0.75
- C0.444
- D0.392
A bag contains 4 green and 6 yellow counters. Two counters are taken at random without replacement. Find the probability that they are different colours.
- A\(\dfrac{8}{15}\)
- B\(\dfrac{12}{25}\)
- C\(\dfrac{4}{15}\)
- D\(\dfrac{7}{15}\)
A box contains 5 red, 4 blue and 3 white pens. Four pens are taken at random. Find the probability that exactly 2 of them are red.
- A\(\dfrac{14}{33}\)
- B0.354
- C\(\dfrac{7}{99}\)
- D\(\dfrac{2}{99}\)
Mia has up to three attempts to pass a test and stops as soon as she passes. The probability that she passes is 0.5 on her first attempt, 0.4 on her second and 0.3 on her third. Find the probability that she passes.
- A0.79
- B0.21
- C0.94
- D0.7
The 7 letters of the word BANANAS are arranged in a random order. Find the probability that the three As are next to each other.
- A\(\dfrac{1}{7}\)
- B\(\dfrac{2}{7}\)
- C\(\dfrac{6}{7}\)
- D\(\dfrac{1}{35}\)
Two fair dice are thrown. Find the probability that at least one die shows a 6 or the total is 8.
- A\(\dfrac{7}{18}\)
- B\(\dfrac{4}{9}\)
- C\(\dfrac{1}{18}\)
- D\(\dfrac{11}{36}\)
Quiz key
Total 2: (1,1); total 3: (1,2), (2,1); total 4: (1,3), (2,2), (3,1). That is 6 of the 36 equally likely outcomes, so \(\dfrac{6}{36} = \dfrac{1}{6}\). The tempting \(\dfrac{3}{11}\) treats the 11 totals 2 to 12 as equally likely, which they are not. \(\dfrac{1}{9}\) counts unordered pairs, missing that (1,2) and (2,1) are different outcomes, and \(\dfrac{5}{18}\) includes the total 5.
Independent, so \(\text{P}(A \cap B) = 0.4 \times 0.35 = 0.14\). Then \(\text{P}(A \cup B) = 0.4 + 0.35 - 0.14 = 0.61\). Or: neither happens with probability \(0.6 \times 0.65 = 0.39\), and \(1 - 0.39 = 0.61\). The tempting 0.75 adds as if the events were exclusive. 0.14 is \(\text{P}(A \cap B)\), and 0.39 is the probability that neither happens.
No bulbs fail: \(0.85^5 = 0.4437\). At least one fails: \(1 - 0.4437 = 0.556\) (3 s.f.). The tempting 0.75 adds 0.15 five times, but the failures are not exclusive (two bulbs can both fail). 0.444 is the probability that none fail, and 0.392 is exactly one failure, \(5 \times 0.15 \times 0.85^4\).
GY: \(\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}\). YG: \(\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\). Total \(\dfrac{48}{90} = \dfrac{8}{15}\). The tempting \(\dfrac{12}{25}\) keeps 10 counters for the second draw, as if with replacement. \(\dfrac{4}{15}\) counts only one order, and \(\dfrac{7}{15}\) is the probability of the same colour.
Total \({}^{12}C_{4} = 495\). Favourable \({}^{5}C_{2} \times {}^{7}C_{2} = 10 \times 21 = 210\). \(\dfrac{210}{495} = \dfrac{14}{33}\). The tempting 0.354 is \(6\left(\tfrac{5}{12}\right)^2\left(\tfrac{7}{12}\right)^2\), which assumes the pens are replaced. \(\dfrac{7}{99}\) is the single order RRNN, forgetting the other 5 orders, and \(\dfrac{2}{99}\) chooses the red pens but not the other two.
Pass first time: 0.5. Fail then pass: \(0.5 \times 0.4 = 0.2\). Fail twice then pass: \(0.5 \times 0.6 \times 0.3 = 0.09\). Total \(0.79\). Check: she fails all three with probability \(0.5 \times 0.6 \times 0.7 = 0.21\), and \(1 - 0.21 = 0.79\). 0.21 is that failure probability. The tempting 0.94 is \(1 - 0.5 \times 0.4 \times 0.3\), which multiplies the pass probabilities instead of the fail probabilities, and 0.7 forgets the third attempt.
All arrangements: \(\dfrac{7!}{3!\,2!} = 420\). With AAA as one unit, the units B, AAA, N, N, S give \(\dfrac{5!}{2!} = 60\). Probability \(\dfrac{60}{420} = \dfrac{1}{7}\). The tempting \(\dfrac{2}{7}\) uses \(5! = 120\), forgetting the two identical Ns. \(\dfrac{6}{7}\) multiplies by \(3!\) for the As, which are identical, and \(\dfrac{1}{35}\) is the probability that the As fill the first three places only.
At least one 6: 11 outcomes. Total 8: (2,6), (3,5), (4,4), (5,3), (6,2), 5 outcomes. Both: (2,6) and (6,2), 2 outcomes. Either: \(11 + 5 - 2 = 14\), so \(\dfrac{14}{36} = \dfrac{7}{18}\). The tempting \(\dfrac{4}{9} = \dfrac{16}{36}\) adds the two events and counts (2,6) and (6,2) twice. \(\dfrac{1}{18}\) is the overlap only, and \(\dfrac{11}{36}\) is the first event only.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.