At a glance. One lesson of about 60 minutes. It follows conditional probability and closes the unit. Students use natural frequencies and a tree to find the share of positives who have a condition, find the total probability of a positive result, solve for a prevalence, and write a justified decision.
Calculator: no calculator. No calculator is needed. Rates are chosen so that every count is a whole number, and the thinking is the structure of the tree and the choice of the group.
What students will learn
Know
- P(positive) is the total of the routes on the tree that end in a positive.
- P(has it | positive) = (route through has it and positive) ÷ P(positive).
- Sensitivity, the false alarm rate and prevalence are all needed to interpret a positive result.
Understand
- A conditional probability reversed can be very different from the original.
- How common a condition is changes how much a positive result tells us.
- A good decision uses the numbers, names what they do and do not show, and says what should happen next.
Do
- Compute a share of positives from counts out of a large group.
- Compute P(A | B) from P(A), P(B | A) and P(B | A′).
- Write a short, justified decision about screening.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: using counts and trees to find the total and the reversed conditional probability (questions 1 to 7).
- Criterion B. Investigating patterns: the figure and the what-if. Students find how the share of true positives changes with how common the condition is and with a second test.
- Criterion C. Communicating: tree diagrams and counts laid out in an ordered way, and a decision written with its reasons (worked example 3).
- Criterion D. Applying mathematics in real-life contexts: the screening hook and the exit ticket, which ask for a justified health decision.
The one thing to do with the figure
Ask for a prediction before every move. Before anyone reads the readout, ask: “What share of the positives do you expect to have it?” Then read it together and ask why it is so far from the 90% catch rate.
Then set the condition to 1% and the false alarm rate to 8%, and ask: “Why did it drop?” Students should say that the healthy group is large and even a small false alarm rate gives many false positives.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the 98% test. Collect predictions of how many positives have it. |
| 12 min | The figure with predictions; the key idea; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose A or D, ask which group each fraction divides by. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | 0.3; 20; 0.12. |
|---|---|
| Hook | About 6 in 100: of 500000 people, 1000 have it and 980 test positive, but 3% of 499000 is about 14970 false alarms, so 980 of about 15950, about 6%. |
| Worked examples | About 31% (34 of 110); about 0.27; no, because about 3 in 4 positives are false alarms, so follow up with a second test. |
| Faded example | 24 of 121, about 0.2. |
| 1. Which fraction is the reversed conditional | B. A is the test rate; C divides by everyone; D is the false-alarm share. |
| 2. Counting positives | 67. 40 + 27. |
| 3. Share of positives who have it | 0.75. 120 ÷ 160. |
| 4. Total probability from a tree | 0.25. 0.07 + 0.18. |
| 5. Reversing a conditional | 0.28. 0.07 ÷ 0.25. |
| 6. How common must it be | 25. 0.9p = 0.3(1 − p). |
| 7. False alarms per true positive | 4.4. 396 ÷ 90. |
| 8. Justifying a screening decision | C. A has the right share but the wrong decision; B gives the test rate; D uses the false alarm rate. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, C or D in question 1 | Gave the test rate, divided by everyone, or gave the share of false alarms. | Ask which group the person is in once the test is positive. |
| 40 or 70 in question 2 | Counted only the true positives, or took the false alarm rate of everyone. | Split into with and without first. |
| 0.6 or 0.12 in question 3 | Gave the test rate or divided by everyone. | Count positives in each group, then divide. |
| 0.07 or 0.9 in question 4 | Used one route, or added the rates. | Weight each rate by the size of its group. |
| 0.7, 0.07 or 0.35 in question 5 | Gave the test rate, one route, or divided by the false alarm rate. | Write the fraction as route over total. |
| 50, 30 or 75 in question 6 | Gave the target share, the false alarm rate, or the other group. | Write two expressions in p and set them equal. |
| 0.23 or 3.96 in question 7 | Inverted the ratio, or divided by the people with it. | Find 90 and 396 first, then divide the second by the first. |
| A, B or D in question 8 | Chose treatment despite the share, quoted the test rate, or used the false alarm rate as the share. | Do the 4000 students as counts: 20 with it, 18 positive, 199 false alarms. |
Questions worth asking
- Why does a rare condition give many false positives? Most people are healthy, so even a small false alarm rate on a large group gives many false alarms.
- Does a test with fewer false alarms always help? Yes, it raises the share of positives that are real, but the prevalence still matters.
- What should a screening programme do with a positive result? Treat it as a reason for a second, more accurate test rather than as a diagnosis.
- How is this different from P(B | A) = P(B)? Here the test and the condition are strongly dependent, but the dependence is imperfect, which is the point.
Differentiation
A way in for anyone stuck
- Give a table with four boxes: has it and positive, has it and negative, does not and positive, does not and negative, for students to fill with counts out of 10000.
- Provide a tree frame with the first branches labelled and the rates written at the ends of the second branches.
- Do the first question with counts out of 100 before using thousands.
Stretch
- Write Bayes' theorem P(A | B) = P(B | A) P(A) ÷ [P(B | A) P(A) + P(B | A′) P(A′)] and verify it with question 5.
- Find the false alarm rate needed for a test that catches 95% of cases in a condition affecting 1 in 200, so that 3 in 4 positives are real.
Students learning in English as an additional language
“Positive” here means that the test result says the condition is present. “False alarm” or “false positive” means a positive result for a person who does not have it. “Prevalence” means how common the condition is. “Screening” means testing people who have no symptoms.
| Word | Plain meaning | Example |
|---|---|---|
| sensitivity | the share of people with the condition whom the test catches | 90 of every 100 people with it test positive |
| false positive | a positive result for a person who does not have the condition | a healthy student told the test is positive |
| prevalence | the share of people who have the condition | 1 person in 500 |
| screening | testing people who have no symptoms to find a condition early | a school health check offered to everyone |
| natural frequencies | counts out of a large group instead of probabilities | out of 1000 people, 30 have it |
| total probability | the sum over every route of a tree that gives an outcome | P(positive) = true positive route + false positive route |
Sentence starters for explaining aloud or in writing:
- Out of … people, … have it and … test positive.
- Of the positives, … have the condition, so a positive result means …
- I recommend … because …
Two things not to say
“The test is 98% accurate, so a positive means 98%.” Accuracy of a test is not the chance a positive is right.
“Screening is pointless when it gives false alarms.” A positive result can still be strong evidence that justifies a second test.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- CountOf 1000 people, 10 have a condition, and the test catches 8 of them. How many test positive correctly?
8. - Earlier work (lesson 5)P(A ∩ B) = 0.12 and P(B) = 0.4. Find P(A | B).
0.3.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA condition affects 4% of a group. A test catches 75% of cases and gives a false alarm to 10% of the others. Out of 1000 people, how many test positive?
30 + 96 = 126. - TreeP(has it) = 0.3, P(positive | has it) = 0.8 and P(positive | does not) = 0.1. Find P(has it | positive).
0.24 ÷ (0.24 + 0.07) = 24/31, about 0.77.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningExplain why making a test catch more cases (higher sensitivity) usually helps less than lowering the false alarm rate for a rare condition.
The healthy group is large, so most positives are false alarms; a lower false alarm rate cuts them directly, while a higher sensitivity only adds to the small group of true positives. - DecisionA school will test 400 students for a condition affecting 1 in 40, using a test that catches 80% and has a 5% false alarm rate. Write a recommendation.
10 have it, 8 test positive; 390 do not, about 20 false alarms. About 8 of 28 positives, 29%, have it. Follow every positive with a second test; do not treat or label students on one result.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No calculator is needed. Keep contexts general: do not ask students to disclose their own health information.