The shape of the lesson
By the end of the lesson, learners can
- Given that: In the lab's conditional tab, click on a condition and watch the sample space shrink to just those outcomes.
- Reversing a tree: Trees are drawn in time order, so their second branches are conditional probabilities such as P(late | rain).
- Testing independence: To decide whether two events are independent, calculate \(\text{P}(A \cap B)\) and \(\text{P}(A) \times \text{P}(B)\) separately and compare.
- Exclusive is not independent: If \(A\) and \(B\) are exclusive, knowing \(B\) happened tells you for certain that \(A\) did not, which is the opposite of independence.
- Formula: \(\text{P}(A \mid B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)}\).
- Tables: the condition picks the row or column total to divide by.
- Trees: find \(\text{P}(B)\) by adding every route through \(B\), then divide.
- Independence: calculate \(\text{P}(A \cap B)\) and \(\text{P}(A)\,\text{P}(B)\); state equal or not.
- Exclusive: \(\text{P}(A \cap B) = 0\), so (with non-zero probabilities) not independent.
One picture every idea returns to
\(\text{P}(A \mid B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)}\): the probability of \(A\) once you know \(B\) has happened. Events are independent exactly when \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\), which is the same as \(\text{P}(A \mid B) = \text{P}(A)\).
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, the odds shift.. Knowing that one event has happened shrinks the sample space. 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: Given that | In the lab's conditional tab, click on a condition and watch the sample space shrink to just those outcomes. Project tab 4 of the lab (Combined probability lab, conditional probability) and let a student drive it while the class predicts. Model the worked example From a two-way table on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Reversing a tree | Trees are drawn in time order, so their second branches are conditional probabilities such as P(late | rain). Model the worked example Was it raining? on the board, one line at a time. Your turn: A bag has 5 red and 3 blue balls. Two are taken without replacement. Given that the second ball is red, find the probability that the first was red. | Pairs |
| 17 to 23 min | Teach and check: Testing independence | To decide whether two events are independent, calculate \(\text{P}(A \cap B)\) and \(\text{P}(A) \times \text{P}(B)\) separately and compare. Project tab 3 of the lab (Combined probability lab, Venn diagrams) and let a student drive it while the class predicts. Model the worked example Two dice on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Exclusive is not independent | If \(A\) and \(B\) are exclusive, knowing \(B\) happened tells you for certain that \(A\) did not, which is the opposite of independence. Model the worked example Both ways on the board, one line at a time. | Jigsaw |
| 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, 25 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 Exclusive is not independent, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 |
|---|---|
| Mixing up \(\text{P}(A \mid B)\) and \(\text{P}(B \mid A)\). | They have the same numerator but different denominators: \(\tfrac{1}{2}\) and \(\tfrac{5}{8}\) above. |
| Answering 0.4. | That is P(late | rain), read straight off the branch; the question asks for P(rain | late), which needs the whole calculation. |
| Deciding independence by intuition ('the dice are separate, so the events must be independent'). | Events defined on the same throw can go either way: \(A\) is independent of \(B\) but not of \(C\). |
| Writing 'independent, because they cannot happen together'. | That describes exclusive events, and it shows the events are NOT independent. |
Conditional probability and independence: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
\(\text{P}(A) = 0.5\), \(\text{P}(B) = 0.6\) and \(\text{P}(A \cap B) = 0.2\). Find \(\text{P}(A \mid B)\) and \(\text{P}(B \mid A)\), and determine whether \(A\) and \(B\) are independent.
200 students were asked how they travel to college.
| Car | Bus | Walk | Total | |
|---|---|---|---|---|
| Year 12 | 30 | 45 | 25 | 100 |
| Year 13 | 40 | 35 | 25 | 100 |
| Total | 70 | 80 | 50 | 200 |
A student is chosen at random. (a) Find P(bus | Year 12). (b) Find P(Year 13 | car). (c) Determine whether the events 'walks' and 'Year 12' are independent.
2% of a population have a condition. A test gives a positive result for 95% of people with the condition and for 8% of people without it. A person is chosen at random and tested.
(a) Find the probability that the result is positive. (b) Given that the result is positive, find the probability that the person has the condition.
Two fair dice are thrown. \(A\) is 'the first die shows 3', \(B\) is 'the total is 7' and \(C\) is 'the total is 9'. Determine whether \(A\) and \(B\) are independent, and whether \(A\) and \(C\) are independent.
A box contains 4 red and 6 blue pens. Two pens are taken at random without replacement.
(a) Find the probability that the second pen is blue. (b) Given that the second pen is blue, find the probability that the first pen was red.
A factory has two machines. Machine A makes 60% of the items and machine B makes the rest. 3% of items from A and 5% of items from B are faulty.
(a) An item is chosen at random. Find the probability that it is faulty. (b) Given that an item is faulty, find the probability that it was made by machine A. (c) Two items are chosen at random, independently, and both are faulty. Find the probability that both were made by machine B.
Answer key and marking notes
- \(\text{P}(A \mid B) = \tfrac{0.2}{0.6} = \tfrac{1}{3}\); \(\text{P}(B \mid A) = \tfrac{0.2}{0.5} = 0.4\).
- \(\text{P}(A)\,\text{P}(B) = 0.3 \ne 0.2\), so not independent.
- (a) \(\tfrac{45}{100} = 0.45\).
- (b) \(\tfrac{40}{70} = \tfrac{4}{7}\).
- (c) P(walk and Year 12) \(= \tfrac{25}{200} = 0.125\); P(walk) \(\times\) P(Year 12) \(= 0.25 \times 0.5 = 0.125\). Equal, so independent.
- (a) \(0.02 \times 0.95 + 0.98 \times 0.08 = 0.019 + 0.0784 = 0.0974\).
- (b) \(\dfrac{0.019}{0.0974} = 0.195\) (3 s.f.).
- \(\text{P}(A) = \tfrac{1}{6}\), \(\text{P}(B) = \tfrac{6}{36} = \tfrac{1}{6}\), \(\text{P}(C) = \tfrac{4}{36} = \tfrac{1}{9}\).
- \(A \cap B\) is (3,4) only: \(\tfrac{1}{36} = \tfrac{1}{6} \times \tfrac{1}{6}\). Independent.
- \(A \cap C\) is (3,6) only: \(\tfrac{1}{36}\), but \(\tfrac{1}{6} \times \tfrac{1}{9} = \tfrac{1}{54}\). Not independent.
- (a) \(\tfrac{4}{10} \times \tfrac{6}{9} + \tfrac{6}{10} \times \tfrac{5}{9} = \tfrac{24}{90} + \tfrac{30}{90} = \tfrac{54}{90} = \tfrac{3}{5}\).
- (b) \(\dfrac{24/90}{54/90} = \tfrac{24}{54} = \tfrac{4}{9}\).
- (a) \(0.6 \times 0.03 + 0.4 \times 0.05 = 0.018 + 0.02 = 0.038\).
- (b) \(\dfrac{0.018}{0.038} = 0.474\) (3 s.f.).
- (c) For one faulty item, P(B | faulty) \(= \dfrac{0.02}{0.038} = 0.5263\).
- The items are independent, so the probability is \(0.5263^2 = 0.277\) (3 s.f.).
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.
\(\text{P}(A) = 0.45\) and \(\text{P}(A \cap B) = 0.18\). Find \(\text{P}(B \mid A)\).
- A0.4
- B0.081
- C2.5
- D0.27
150 people were asked whether they had been to the cinema in the last month.
| Yes | No | Total | |
|---|---|---|---|
| Under 30 | 36 | 24 | 60 |
| 30 or over | 45 | 45 | 90 |
| Total | 81 | 69 | 150 |
- A\(\dfrac{4}{9}\)
- B\(\dfrac{3}{5}\)
- C\(\dfrac{6}{25}\)
- D\(\dfrac{27}{50}\)
Machine A makes 60% of a factory's bolts and machine B makes the rest. 3% of A's bolts and 5% of B's bolts are faulty. A bolt is chosen at random and found to be faulty. Find the probability that it was made by machine A.
- A0.474
- B0.03
- C0.6
- D0.038
\(\text{P}(A) = 0.3\), \(\text{P}(B) = 0.6\) and \(\text{P}(A \cup B) = 0.72\). Which statement is correct?
- A\(A\) and \(B\) are independent, because \(\text{P}(A \cap B) = 0.18 = \text{P}(A) \times \text{P}(B)\)
- B\(A\) and \(B\) are not independent, because \(\text{P}(A \cup B) \ne \text{P}(A) + \text{P}(B)\)
- C\(A\) and \(B\) are exclusive, because \(\text{P}(A \cup B) < 1\)
- D\(A\) and \(B\) are not independent, because \(\text{P}(A \cap B) \ne 0\)
A bag contains 4 red and 6 blue balls. Two balls are taken at random without replacement. Given that the second ball is blue, find the probability that the first ball was blue.
- A\(\dfrac{5}{9}\)
- B\(\dfrac{3}{5}\)
- C\(\dfrac{1}{3}\)
- D\(\dfrac{2}{3}\)
5% of the people in a town have a certain allergy. A skin test is positive for 90% of people with the allergy and for 6% of people without it. A person chosen at random tests positive. Find the probability that they have the allergy.
- A0.441
- B0.9
- C0.045
- D0.102
Two fair dice are thrown. Event \(A\) is "the first die shows 3". Which of these events is independent of \(A\)?
- AThe total is 7
- BThe total is 9
- CThe total is 4
- DThe second die shows a higher score than the first
\(\text{P}(A) = 0.4\), \(\text{P}(B \mid A) = 0.3\) and \(\text{P}(B \mid A') = 0.5\). Find \(\text{P}(A \mid B')\).
- A0.483
- B0.7
- C0.28
- D0.4
Quiz key
\(\text{P}(B \mid A) = \dfrac{0.18}{0.45} = 0.4\). The tempting 0.081 multiplies instead of dividing. 2.5 divides the wrong way round (a probability cannot exceed 1), and 0.27 subtracts, giving \(\text{P}(A \cap B')\).
Restrict to the 81 people who answered Yes; 36 of them are under 30, so \(\dfrac{36}{81} = \dfrac{4}{9}\). The tempting \(\dfrac{3}{5} = \dfrac{36}{60}\) is P(Yes | under 30), the reverse condition. \(\dfrac{6}{25} = \dfrac{36}{150}\) is P(under 30 and Yes), and \(\dfrac{27}{50}\) is P(Yes).
A and faulty: \(0.6 \times 0.03 = 0.018\). B and faulty: \(0.4 \times 0.05 = 0.02\). P(faulty) \(= 0.038\). P(A | faulty) \(= \dfrac{0.018}{0.038} = \dfrac{9}{19} = 0.474\) (3 s.f.). The tempting 0.03 is P(faulty | A), read straight off the branch. 0.6 ignores the information that the bolt is faulty, and 0.038 is P(faulty).
\(\text{P}(A \cap B) = 0.3 + 0.6 - 0.72 = 0.18\), and \(\text{P}(A) \times \text{P}(B) = 0.18\). Equal, so \(A\) and \(B\) are independent. The tempting second statement only shows the events are not exclusive, which says nothing against independence. A non-zero \(\text{P}(A \cap B)\) also shows they are not exclusive, and a union less than 1 proves nothing.
P(BB) \(= \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90}\); P(RB) \(= \dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}\). P(second blue) \(= \dfrac{54}{90}\). P(first blue | second blue) \(= \dfrac{30}{54} = \dfrac{5}{9}\). The tempting \(\dfrac{3}{5}\) ignores the condition and gives P(first blue). \(\dfrac{1}{3}\) is P(BB), not divided by P(second blue), and \(\dfrac{2}{3}\) is P(second blue | first red).
Allergy and positive: \(0.05 \times 0.9 = 0.045\). No allergy and positive: \(0.95 \times 0.06 = 0.057\). P(positive) \(= 0.102\). P(allergy | positive) \(= \dfrac{0.045}{0.102} = 0.441\) (3 s.f.). The tempting 0.9 is P(positive | allergy), the reverse condition. 0.045 is P(allergy and positive), and 0.102 is P(positive).
\(\text{P}(A) = \tfrac{1}{6}\). Total 7: \(\text{P} = \tfrac{6}{36}\), and \(A \cap E\) is (3,4) only, \(\tfrac{1}{36} = \tfrac{1}{6} \times \tfrac{1}{6}\): independent. Total 9: \(\tfrac{1}{36}\) against \(\tfrac{1}{6} \times \tfrac{4}{36} = \tfrac{1}{54}\). Total 4: \(\tfrac{1}{36}\) against \(\tfrac{1}{72}\). Second higher: (3,4), (3,5), (3,6) give \(\tfrac{3}{36}\) against \(\tfrac{1}{6} \times \tfrac{15}{36} = \tfrac{5}{72}\). The tempting idea that every total is independent of one die fails: only 7 can be reached from every first score, so only total 7 works.
\(\text{P}(A \cap B') = 0.4 \times 0.7 = 0.28\). \(\text{P}(A' \cap B') = 0.6 \times 0.5 = 0.3\). \(\text{P}(B') = 0.58\). \(\text{P}(A \mid B') = \dfrac{0.28}{0.58} = 0.483\) (3 s.f.). The tempting 0.7 is \(\text{P}(B' \mid A)\), the reverse condition. 0.28 is \(\text{P}(A \cap B')\) not yet divided, and 0.4 is \(\text{P}(A)\), which would only be right if the events were independent.
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.