The shape of the lesson
By the end of the lesson, learners can
- Shrinking the group: In the lab, pick a 'given that' condition and watch the table: everything outside the chosen row or column fades out.
- Conditions in sample spaces: With two dice, a condition such as 'the total is 8' cuts the 36 cells down to the few that satisfy it.
- Conditional probability on trees: On a tree diagram, the second set of branches is already conditional: each one is a probability GIVEN what happened on the first branch.
- Spot the condition: 'given that', 'if', 'of those who'.
- Restrict: keep only the outcomes in the given group.
- Divide: number in both \(\div\) number in the given group.
- Trees: second branches are already conditional.
- Reverse a tree: wanted path \(\div\) sum of all paths giving the condition.
One picture every idea returns to
'Given that B' means look ONLY at the outcomes in B. Then \(\text{P(A given B)} = \dfrac{\text{number in both A and B}}{\text{number in B}}\). The denominator is the size of the given group, not the whole total.
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, Given that.... Extended. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Shrinking the group | In the lab, pick a 'given that' condition and watch the table: everything outside the chosen row or column fades out. Project tab 4 of the lab (Combined probability lab, conditional) and let a student drive it while the class predicts. Model the worked example From a Venn diagram on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Conditions in sample spaces | With two dice, a condition such as 'the total is 8' cuts the 36 cells down to the few that satisfy it. Model the worked example Given the total on the board, one line at a time. Your turn: Two fair dice are rolled. Given that the total is 10, find the probability that at least one dice shows a 5. | Pairs |
| 21 to 30 min | Teach and check: Conditional probability on trees | On a tree diagram, the second set of branches is already conditional: each one is a probability GIVEN what happened on the first branch. Model the worked example Without replacement on the board, one line at a time. | Round Table |
| 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, 18 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 Conditional probability on trees, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| 'P(A given B)' and 'P(B given A)' are different. | In Fig. 1 they are \(\tfrac{4}{9}\) and \(\tfrac{2}{5}\). Read which group is GIVEN: that group is the denominator. |
| Do not divide by 36 once a condition is given. 'Given that the total is 8' means the denominator is 5, not 36. | Return to Conditions in sample spaces: ask a pair to state the correct version and show it on the lab before moving on. |
| The branch value 0.4 is P(late GIVEN rain), not P(rain GIVEN late). | Reversing the condition needs the division by P(late). |
Conditional probability: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
In a Venn diagram of 30 students, 6 do chess only, 4 do chess and art, 11 do art only and 9 do neither.
Find the probability that a student does art, given that they do chess.
200 people were asked which phone brand they use.
| Brand A | Brand B | |
|---|---|---|
| Under 30 | 48 | 32 |
| 30 or over | 54 | 66 |
Find (a) P(brand A, given under 30) (b) P(30 or over, given brand B).
A bag holds 6 red and 4 blue counters. Two are taken without replacement.
(a) Given that the first counter is blue, find the probability that the second is blue. (b) Find the probability that both are blue.
Two fair dice are rolled. Given that at least one of the dice shows a 6, find the probability that the total is 10 or more.
Machine A makes 60% of the items in a factory and machine B makes the rest. 3% of the items from A are faulty and 5% of the items from B are faulty.
(a) Find the probability that an item chosen at random is faulty. (b) Given that an item is faulty, find the probability that it was made by machine B.
In a school, the probability that a student studies History is 0.5, Geography is 0.4, and both is 0.15.
Find the probability that a student (a) studies Geography, given that they study History (b) studies History, given that they do NOT study Geography.
Answer key and marking notes
- Chess group: \(6 + 4 = 10\) students.
- 4 of them do art: \(\tfrac{4}{10} = \tfrac{2}{5}\)
- (a) Under 30: 80 people, 48 use A: \(\tfrac{48}{80} = \tfrac{3}{5}\)
- (b) Brand B: \(32 + 66 = 98\) people, 66 are 30 or over: \(\tfrac{66}{98} = \tfrac{33}{49}\)
- (a) 9 left, 3 blue: \(\tfrac{3}{9} = \tfrac{1}{3}\)
- (b) \(\tfrac{4}{10} \times \tfrac{1}{3} = \tfrac{4}{30} = \tfrac{2}{15}\)
- Cells with at least one 6: \(6 + 6 - 1 = 11\).
- Totals of 10 or more among them: (4,6), (6,4), (5,6), (6,5), (6,6), so 5.
- Probability \(\tfrac{5}{11}\)
- Paths: A and faulty \(0.6 \times 0.03 = 0.018\); B and faulty \(0.4 \times 0.05 = 0.02\).
- (a) \(0.018 + 0.02 = 0.038\)
- (b) \(\dfrac{0.02}{0.038} = 0.526\) (3 s.f.)
- Venn regions: History only 0.35, both 0.15, Geography only 0.25, neither 0.25.
- (a) \(\dfrac{0.15}{0.5} = 0.3\)
- (b) Not Geography: \(1 - 0.4 = 0.6\); of this, History only is 0.35.
- \(\dfrac{0.35}{0.6} = \dfrac{7}{12}\) (0.583)
6 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.
The table shows the drink chosen by 90 people at a cafe.
| Tea | Coffee | Total | |
|---|---|---|---|
| Adults | 24 | 36 | 60 |
| Teenagers | 18 | 12 | 30 |
| Total | 42 | 48 | 90 |
- A\(\tfrac{3}{7}\)
- B\(\tfrac{3}{5}\)
- C\(\tfrac{1}{5}\)
- D\(\tfrac{7}{15}\)
In a group of 60 people, 20 are in set P only, 9 are in both P and Q, 15 are in Q only and 16 are in neither. A person is chosen at random. Given that the person is in P, find the probability that they are also in Q.
- A\(\tfrac{9}{29}\)
- B\(\tfrac{3}{8}\)
- C\(\tfrac{3}{20}\)
- D\(\tfrac{9}{20}\)
Two fair six-sided dice are rolled. Given that at least one dice shows a 4, find the probability that the total is even.
- A\(\tfrac{5}{11}\)
- B\(\tfrac{5}{36}\)
- C\(\tfrac{1}{2}\)
- D\(\tfrac{5}{12}\)
A bag contains 7 red and 5 blue sweets. Two sweets are taken at random without replacement. Given that the two sweets are the same colour, find the probability that they are both red.
- A\(\tfrac{21}{31}\)
- B\(\tfrac{7}{22}\)
- C\(\tfrac{49}{74}\)
- D\(\tfrac{7}{12}\)
A shop sells phones from two factories. 60% come from factory A and 40% from factory B. Of the phones from A, 5% are faulty; of those from B, 8% are faulty. A phone is chosen at random and is faulty. Find the probability that it came from factory B, correct to 3 significant figures.
- A0.516
- B0.08
- C0.032
- D0.4
For two events, \(\text{P}(A) = 0.5\), \(\text{P}(B) = 0.4\) and \(\text{P}(A \cap B) = 0.15\). Find the probability of \(A\), given that \(B\) does not happen.
- A0.583
- B0.35
- C0.375
- D0.7
Quiz key
The given group is the 42 tea drinkers, and 18 of them are teenagers: \(\tfrac{18}{42} = \tfrac{3}{7}\). The tempting \(\tfrac{3}{5}\) is \(\tfrac{18}{30}\), the probability of tea GIVEN teenager, the reverse condition. \(\tfrac{1}{5}\) divides by all 90 people, and \(\tfrac{7}{15}\) is P(tea).
Circle P holds \(20 + 9 = 29\) people, and 9 of them are also in Q: \(\tfrac{9}{29}\). The tempting \(\tfrac{3}{8}\) is \(\tfrac{9}{24}\), the probability of P given Q. \(\tfrac{3}{20}\) divides by all 60, and \(\tfrac{9}{20}\) divides by the P-only region instead of the whole circle.
Cells with at least one 4: \(6 + 6 - 1 = 11\) ((4, 4) is counted once). Even totals among them: (4, 2), (4, 4), (4, 6), (2, 4), (6, 4), which is 5 cells, so \(\tfrac{5}{11}\). The tempting \(\tfrac{5}{36}\) divides by all 36 cells and ignores the condition. \(\tfrac{1}{2}\) is P(even total) with no condition at all, and \(\tfrac{5}{12}\) counts (4, 4) twice in the given group.
P(RR) \(= \tfrac{7}{12} \times \tfrac{6}{11} = \tfrac{42}{132}\); P(BB) \(= \tfrac{5}{12} \times \tfrac{4}{11} = \tfrac{20}{132}\). P(same colour) \(= \tfrac{62}{132}\). So P(RR given same) \(= \tfrac{42}{62} = \tfrac{21}{31}\). The tempting \(\tfrac{7}{22}\) is P(RR) itself, not divided by P(same colour). \(\tfrac{49}{74}\) treats the picks as with replacement, and \(\tfrac{7}{12}\) is just P(first sweet red).
B and faulty: \(0.4 \times 0.08 = 0.032\). A and faulty: \(0.6 \times 0.05 = 0.03\). P(faulty) \(= 0.062\). P(B given faulty) \(= \dfrac{0.032}{0.062} = 0.516\). The tempting 0.08 is P(faulty given B), the condition the wrong way round. 0.032 is P(B and faulty), not divided by P(faulty), and 0.4 ignores the information that the phone is faulty.
A only \(= \text{P}(A \cap B') = 0.5 - 0.15 = 0.35\), and \(\text{P}(B') = 1 - 0.4 = 0.6\). So the answer is \(\dfrac{0.35}{0.6} = 0.583\) (3 s.f.). The tempting 0.35 is \(\text{P}(A \cap B')\), not divided by \(\text{P}(B')\). 0.375 is \(\dfrac{0.15}{0.4}\), the probability of A given B, and 0.7 is \(\dfrac{0.35}{0.5}\), the probability of B' given A.
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.