HSS.MD.B.7Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.MD.B.7: Analyzing Decisions and Strategies with Probability
In plain English: HSS.MD.B.7 is the Common Core statistics and probability standard that asks students to use probability to analyze real decisions and strategies, such as what a positive medical test means, how a product-testing plan performs, or when to pull a hockey goalie. It is an advanced (+) standard, usually taught in a Statistics course or a probability unit in Algebra II or Precalculus.
(+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game).
Common Core State Standards for Mathematics · Domain: Using Probability to Make Decisions (MD) · Cluster: Use probability to evaluate outcomes of decisions Also written as HSS-MD.B.7 or S-MD.7 · Official standard
Students use probability to judge decisions that people and organizations really make. The lesson starts with medical screening: students build a frequency tree and a two-way table to find the chance that a person with a positive result actually has the condition, and see why that chance depends on how common the condition is. They then compare two strategies for testing products before a store accepts a shipment, and analyze why hockey coaches pull the goalie when they trail late in a game.
Across all three contexts students follow the same routine: list the options, assign probabilities to the outcomes, choose a criterion that fits the goal (the chance of a good outcome, an error rate or an expected cost), compute, and then question the model's assumptions before recommending a decision. All data in the lesson are invented but realistic.
Learning Objectives
By the end of this lesson, students will be able to:
Use a frequency tree or two-way table to find P(condition | positive) and P(condition | negative) from the prevalence, sensitivity and specificity of a test
Explain why the same test is more trustworthy in a high-risk group than in a general screening, and evaluate a retesting strategy
Compare product-testing plans by the probability that they accept good lots and bad lots
Choose a decision criterion that fits the goal (probability of winning, error rates or expected cost) and use it to compare strategies such as pulling a hockey goalie
State the assumptions behind a probability model of a decision and explain how they affect the recommendation
Prior Knowledge Required
Students should already be comfortable with:
Two-way frequency tables used as a sample space HSS.CP.A.4
Conditional probability as a fraction of the outcomes in the given event HSS.CP.B.6
The multiplication rule for independent events HSS.CP.A.2
Read the scenario aloud and have every student commit to one of four answers on a sticky note before any discussion.
Warm-Up Prompt
"A screening test catches 90% of people who have a certain condition and wrongly flags 5% of people who do not. About 2% of the people screened have the condition. You test positive. Is the chance that you have the condition closer to 90%, 50%, 25% or 5%?"
Tally the votes on the board without revealing the answer. Many students choose 90% because it is the number in the question. Ask two students with different votes to explain their reasoning, then tell the class that by the end of Direct Instruction they will be able to settle the question with a count of people rather than a guess. Point out that the question is really about a decision: should a person who tests positive start treatment, or get a second test?
Direct Instruction25 minutes
A routine for analyzing a decision. Post these five steps and refer to them in every example:
List the options: for example, treat now or retest; accept the lot or reject it; keep the goalie in or pull the goalie.
List the outcomes and their probabilities: use a frequency tree, a two-way table or the multiplication rule.
Choose a criterion that fits the goal: the probability of a good outcome, an error rate, or an expected value (cost or points).
Compute and compare the criterion for each option.
Question the model: which probabilities were estimated, which events were assumed independent, and would the decision change if they were a little different?
Part 1: Medical testing. Define three terms. Prevalence is the share of the tested group that has the condition. Sensitivity is P(positive | condition). Specificity is P(negative | no condition). Stress that none of these is the number a patient cares about, which is P(condition | positive). Work Example 1 with Diagram 1, then show the same counts as a two-way table:
Example 1 as a two-way table (20,000 people screened)
Test positive
Test negative
Total
Condition
360
40
400
No condition
980
18,620
19,600
Total
1,340
18,660
20,000
Medical testing: reading a positive result
The warm-up test is used on 20,000 people. The prevalence is 2%, the sensitivity is 90% and the specificity is 95%. How likely is a person who tests positive to have the condition?
Equation: 400 have the condition and 360 of them test positive; 19,600 do not and 5% of them, 980, test positive. P(condition | positive) = 360/1,340 ≈ 0.269
Medical testing: a retest strategy
Everyone who tests positive in Example 1 is tested a second time. Assume the two results are independent for a given person. How likely is a person who tests positive twice to have the condition?
Equation: Positive twice: 360(0.90) = 324 with the condition and 980(0.05) = 49 without. P(condition | two positives) = 324/373 ≈ 0.869
Product testing: comparing two plans
A store receives large lots of phone chargers. Plan A tests 5 chargers and accepts the lot only if all 5 pass; Plan B does the same with 10 chargers. Treat each charger as defective independently. Compare the plans for a good lot (2% defective) and a bad lot (10% defective).
Equation: Plan A: P(accept good) = 0.98⁵ ≈ 0.904, P(accept bad) = 0.90⁵ ≈ 0.590. Plan B: 0.98¹⁰ ≈ 0.817 and 0.90¹⁰ ≈ 0.349
Game strategy: pulling the hockey goalie
A team trails by one goal with 2 minutes left. An invented model gives P(tie by the final horn) = 0.09 with the goalie in net and 0.17 with the goalie pulled for an extra skater. Pulling the goalie also raises P(allowing another goal) from 0.10 to 0.40. A tied game goes to overtime, which the team wins half the time.
Equation: Criterion: P(win). Goalie in: 0.09(0.5) = 0.045. Goalie pulled: 0.17(0.5) = 0.085. Pull the goalie: a two-goal loss counts the same as a one-goal loss
Consumer decision: an extended warranty
A $1,200 laptop has a $150 two-year warranty. Without it, the owner estimates a 10% chance of a $400 repair and a separate 2% chance of a $1,200 replacement over two years (assume both cannot happen).
Equation: Expected cost without the warranty = 0.10($400) + 0.02($1,200) = $64, which is less than $150
After Example 1, return to the warm-up votes: the answer is about 27%, closest to 25%. The 980 false positives outnumber the 360 true positives because the healthy group is so much larger. Use Diagram 2 to show that the same test gives P(condition | positive) of 0.818 in a group where 20% have the condition. Example 2 shows why a positive screening result usually leads to a second test rather than to treatment.
Part 2: Strategies. In Example 3 ask students which plan they would choose and why; they should notice that neither plan wins on both lot types, so the store must weigh two kinds of error. In Example 4 the criterion matters: pulling the goalie makes a bigger loss more likely, but the standings do not count goal difference in a loss, so the right criterion is the chance of tying. In Example 5 the expected cost favors skipping the warranty, but ask students when a person might still buy it (a $1,200 loss would be a hardship). Expected value is one criterion, not the only one.
Guided Practice15 minutes
Pairs analyze an email spam filter with invented data for 10,000 emails, 30% of which are spam. At its current setting the filter flags 96% of spam and 2% of real email:
Spam filter at its current setting (10,000 emails)
Flagged
Not flagged
Total
Spam
2,880
120
3,000
Real email
140
6,860
7,000
Total
3,020
6,980
10,000
Ask, one at a time: (1) What is P(real email | flagged)? (140/3,020 ≈ 0.046.) (2) What is P(spam | not flagged)? (120/6,980 ≈ 0.017.) (3) A stricter setting flags 85% of spam and only 0.2% of real email. How many real emails would it flag, and how much spam would get through? (14 real emails flagged; 450 spam emails get through.) (4) The company wants to delete flagged email automatically. Which setting should it use, and what does that choice cost? (The stricter setting: 14 lost real emails per 10,000 instead of 140, at the price of 330 more spam emails in the inbox.) Listen for students who divide by the row total (3,000) instead of the column total (3,020) in question 1: the condition is "flagged", so the flagged column is the sample space.
Independent Practice15 minutes
Students work two problems on their own and write a one-sentence recommendation for each.
Problem A (tennis serves). A player gets two serves per point; missing both loses the point. Her big serve goes in 60% of the time and wins 75% of the points when it goes in. Her safe serve goes in 90% of the time and wins 55% of the points when it goes in. Compare the strategies big-then-safe and big-then-big. (Big-then-safe: 0.60(0.75) + 0.40(0.90)(0.55) = 0.648. Big-then-big: 0.45 + 0.40(0.45) = 0.63. Big-then-safe wins more points.)
Problem B (product testing). A lot is rejected if any of 8 tested items is defective. If 5% of the items in the lot are defective, what is the probability the lot is rejected? (1 - 0.95⁸ ≈ 0.337.) Is this plan strict enough if the store wants to reject such lots most of the time? (No: it rejects them only about a third of the time.)
Circulate and ask each student which criterion they used and why it fits the goal.
Closure5-10 minutes
Exit ticket: (1) A test has 99% sensitivity and 99% specificity, and 1% of the people tested have the condition. Find P(condition | positive). (Answer: 0.5; per 10,000 people, 99 true positives and 99 false positives.) (2) Name one piece of information you would want before deciding whether everyone in a school should take this test, and explain why it matters.
Differentiation Strategies
For Struggling Students
Always start from a round population (1,000, 10,000 or 20,000) and fill in the frequency tree with counts before writing any probability
Give a two-way table template with the labels already filled in and the row and column totals marked, so the conditional probability is read from one row or column
Use one decision template for every context: options, outcomes, probabilities, criterion, recommendation
For Advanced Students
Write P(condition | positive) as a formula in the prevalence p, the sensitivity and the specificity, and use it to find the prevalence at which a test's positive results are right half the time
In Example 3, find how many chargers an all-must-pass plan could test and still accept a good lot at least 90% of the time
Research how an invented goalie model could be estimated from real game data, and list the assumptions such an estimate would need
Assessment Guidance
What to Look For
Check that students can say in words which group a conditional probability is taken from before they compute it: "of the people who test positive" is a different group from "of the people who have the condition". When students compare strategies, ask them to name their criterion and explain why it fits the goal, and look for at least one stated assumption (independence, estimated probabilities, costs). A correct number with no recommendation, or a recommendation with no probability behind it, is only half of what this standard asks.
02
Classroom Activities
3 Activities
1
Who Should Be Screened?
20 minGroups of 3
Each group gets three scenario cards for the same test, used in three different groups of 1,000 people. Students build a frequency tree for each card and use the results to recommend who should be screened.
Scenario Cards
The test has 80% sensitivity and 90% specificity on every card
Card 1: a general screening where 5% of the 1,000 people have the condition (40 true positives, 95 false positives, P(condition | positive) = 40/135 ≈ 0.296)
Card 2: patients with symptoms, 20% have the condition (160 true positives, 80 false positives, 160/240 ≈ 0.667)
Card 3: close contacts of known cases, 50% have the condition (400 true positives, 50 false positives, 400/450 ≈ 0.889)
Procedure
One student builds the tree, one checks the counts add to 1,000, and one writes P(condition | positive) and P(condition | negative); roles rotate for each card
Groups plot their three values of P(condition | positive) on a class chart with prevalence on the horizontal axis
Each group writes a two-sentence policy: which group should get this test, and what should happen after a positive result
Discussion Questions
The test is the same on all three cards. Why does a positive result mean something different on each card?
On Card 1, what would you tell a person who just tested positive?
Which kind of error, a false positive or a false negative, is worse for this condition? Does your answer change your policy?
Challenge Variation
Give Card 1 again with the specificity raised from 90% to 99%. Groups find the new P(condition | positive) and decide whether a more specific test or a more targeted screening does more to reduce false alarms.
2
Pull the Goalie? A Simulation
25 minGroups of 3-4
Groups simulate the last 2 minutes of a hockey game in which their team trails by one goal, using an invented model, and compare the chance of tying the game with the goalie in net and with the goalie pulled.
The Model
Split the 2 minutes into six 20-second segments. In each segment one of three things happens: your team scores, the opponent scores, or nobody scores
Goalie in net: roll two 10-sided dice for a number from 00 to 99. 00-01 means your team scores, 02-03 means the opponent scores, anything else means no goal
Goalie pulled: 00-04 means your team scores, 05-16 means the opponent scores, anything else means no goal
A trial ends when your team ties the game (a success), when the opponent scores (count it as a loss), or after six segments with no goal (a loss)
Procedure
Each group runs 20 trials with the goalie in and 20 with the goalie pulled, recording each trial as T (tie) or L (loss)
Pool the class results and compute the share of ties for each strategy
Compare the pooled results with the exact values from the model: P(tie) = 0.02(1 + 0.96 + 0.96² + ... + 0.96⁵) ≈ 0.109 with the goalie in, and 0.05(1 + 0.83 + 0.83² + ... + 0.83⁵) ≈ 0.198 with the goalie pulled
Discussion Questions
With the goalie pulled, the opponent is much more likely to score. Why is pulling still the better strategy in this model?
Why do small class simulations sometimes point the wrong way, and why does pooling help?
Would you use the same strategy if the game were tied? What would the criterion be then?
Modification for Distance Learning
Use a random number generator set to integers from 0 to 99, or a shared spreadsheet that generates the six segments for each trial. Groups enter their tallies in a shared class table.
3
Design a Quality-Control Plan
20 minPairs
Pairs act as the quality team for a store that buys large lots of earbuds. They must design a testing plan that meets two targets at once and discover that the simplest kind of plan cannot do it.
The Targets
A good lot (2% defective) must be accepted at least 90% of the time
A bad lot (10% defective) must be accepted at most 40% of the time
Treat each tested item as defective independently, with the lot's defect rate
Procedure
Step 1: try plans that accept only if all n tested items pass. Pairs find that 0.98ⁿ ≥ 0.90 needs n ≤ 5, while 0.90ⁿ ≤ 0.40 needs n ≥ 9, so no all-must-pass plan meets both targets
Step 2 (challenge): try plans that accept if at most 1 of the n items is defective, using P(accept) = (1 - d)ⁿ + n·d(1 - d)ⁿ⁻¹ for defect rate d
Pairs find that n = 20 works (about 0.940 for good lots and 0.392 for bad lots) and that n = 19 fails the second target (about 0.420)
Discussion Questions
Testing an earbud destroys it. What does the plan with n = 20 cost compared with a plan that tests 5?
Which error hurts the store more: rejecting a good lot or accepting a bad one? How would that change the targets?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Frequency Tree for a Screening Test
Example 1 as counts: 20,000 people, 2% prevalence, 90% sensitivity and 95% specificity. The two shaded boxes are all the positive results. Only 360 of the 1,340 positives come from people who have the condition, so a positive result means about a 27% chance of having it.
Diagram 2: How Prevalence Changes What a Positive Result Means
P(condition | positive) for the same test (90% sensitivity, 95% specificity) as the prevalence in the tested group goes from 0% to 50%, drawn to scale. In a general screening with 2% prevalence a positive result is right about 27% of the time; in a high-risk group with 20% prevalence it is right about 82% of the time.
04
Homework Assignment
~30 min
HSS.MD.B.7 Homework: Probability and Decisions
Directions: Show a frequency tree, a table or the probability calculation for every answer, and round probabilities to three decimal places. End each problem with a recommendation in one or two sentences that names your criterion. All data are invented.
Part 1: Medical Testing (Problems 1-2)
A clinic gives a rapid strep test to 1,000 children with sore throats; 30% of them actually have strep throat. The test has 85% sensitivity and 95% specificity. (a) Build a frequency tree. (b) Find P(strep | positive) and P(strep | negative). (c) Explain why a doctor might treat after a positive rapid test but send a throat culture to a lab after a negative one.
A screening test for a rare condition has 98% sensitivity and 97% specificity. (a) In a general screening of 100,000 people with 0.5% prevalence, find P(condition | positive). (b) Find P(condition | positive) in a high-risk group where 10% have the condition. (c) A health department can afford to screen only one of these groups. Which would you recommend, and why?
Part 2: Product Testing (Problems 3-4)
A bike-helmet maker crash-tests 6 helmets from each large lot and ships the lot only if all 6 pass; the tested helmets are destroyed. (a) Find the probability that a lot is shipped if 1% of its helmets are defective, and if 8% are defective. (b) Repeat part (a) for a plan that tests 15 helmets. (c) Which plan would you recommend? Discuss both kinds of error and the cost of destroyed helmets.
A factory's items are 4% defective. An inspection catches 90% of defective items but also wrongly rejects 3% of good items. The factory is considering a second, independent inspection; an item is shipped only if it passes both. (a) For one inspection and for two, find the probability that a random item is a shipped defective and the probability that it is a rejected good item. (b) A shipped defective costs $500 and a rejected good item costs $20. Find the expected cost per item for each option. (c) If the second inspection costs $1.50 per item, which option should the factory choose?
Part 3: Game Strategies (Problems 5-6)
A basketball team trails by 2 points with time for one last shot. Their 2-point shot goes in 50% of the time, and if the game goes to overtime they win half the time. Their 3-point shot goes in 35% of the time. (a) Find P(win) for each choice. (b) Which shot should they take? (c) For what 3-point percentage would the two choices give the same P(win)?
A hockey team trails by one goal with 2 minutes left. An invented model splits the time into four 30-second segments. With the goalie in net, in each segment the team scores with probability 0.03 and the opponent scores with probability 0.03. With the goalie pulled, the team scores with probability 0.07 and the opponent with probability 0.15. The team ties only if it scores before the opponent does. (a) Use a tree to find P(tie) for each strategy. (b) Which strategy do you recommend? (c) Name one assumption of this model that a real coach might question.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Model and Setup
Correct tree, table or product for every option, with counts or probabilities labeled
Setup mostly correct, one branch or count wrong
No model or an incorrect one
Conditional Probabilities
Conditions read from the correct group and computed correctly
Correct method with an arithmetic error
Condition reversed or missing
Criterion and Recommendation
Criterion named, fits the goal, and the recommendation follows from it
Recommendation given but the criterion is unclear
No recommendation
Assumptions and Context
At least one assumption or cost discussed and its effect explained
Assumption named but not explained
No discussion of the model
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. All data are invented. Round probabilities to three decimal places.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A screening test has a sensitivity of 92%. Which statement says the same thing?
Answer: A
Sensitivity is P(positive | condition): the share of people with the condition whom the test catches. Choice B reverses the condition; that is P(condition | positive), which also depends on the prevalence. Choice D describes specificity, and choice C mixes both groups together.
Question 2 of 20 · Multiple Choice
A test has a specificity of 97%. What share of the people without the condition get a false positive result?
Answer: C
Specificity is P(negative | no condition) = 0.97, so P(positive | no condition) = 1 - 0.97 = 0.03, or 3%. Choice A gives the share who test negative. Choice D is wrong because this probability is conditional on having no condition; the prevalence is needed only for questions about all people or about all positives.
Question 3 of 20 · Multiple Choice
In a town of 5,000 people, 4% have a condition. A test has 75% sensitivity and 90% specificity. How many false positives should the town expect if everyone is tested?
Answer: D
5,000 - 200 = 4,800 people do not have the condition, and 10% of them test positive: 480 false positives. Choice A is the number of true positives (75% of 200). Choice B is the number of missed cases (25% of 200). Choice C takes 10% of the whole town, including the people who have the condition.
Question 4 of 20 · Multiple Choice
Use the town in the previous question. If a resident tests positive, what is the probability that the resident has the condition?
Answer: B
There are 150 true positives and 480 false positives, so P(condition | positive) = 150/630 ≈ 0.238. Choice A is the sensitivity, the probability in the reverse direction. Choice D divides 150 by 480, leaving the true positives out of the total. Choice C divides by the whole town instead of by the people who tested positive.
Question 5 of 20 · Multiple Choice
A test is moved from a general screening, where 2% of people have the condition, to a clinic where 30% of the patients have it. The sensitivity and specificity do not change. What happens to P(condition | positive)?
Answer: D
With more people who have the condition and fewer who do not, true positives grow and false positives shrink, so a larger share of positives are real. Choice A is a common error: sensitivity and specificity describe the test, but P(condition | positive) also depends on the group tested.
Question 6 of 20 · Multiple Choice
After one positive result, a patient's probability of having a condition is 0.25. The patient takes a second test with 80% sensitivity and 90% specificity, and the two results are independent for a given person. The second result is also positive. What is the probability now?
Answer: A
Start with 1,000 such patients: 250 have the condition and 750 do not. Positive again: 250(0.80) = 200 and 750(0.10) = 75. P(condition) = 200/275 ≈ 0.727. Choice B is the sensitivity. Choice D ignores the new evidence.
Question 7 of 20 · Multiple Choice
A store accepts a large lot of batteries only if all 4 tested batteries pass. If 5% of the batteries in the lot are defective, what is the probability that the store accepts the lot?
Answer: B
Each battery passes with probability 0.95, and the tests are treated as independent, so P(accept) = 0.95⁴ ≈ 0.815. Choice A multiplies 4 by 0.05. Choice C is the probability for a single battery. Choice D is the probability that the lot is rejected.
Question 8 of 20 · Multiple Choice
A lot is accepted only if every tested item passes. What happens when the number of items tested is increased?
Answer: C
P(accept) = (1 - d)ⁿ gets smaller as n grows for every defect rate d greater than 0, so bad lots are caught more often, but good lots are also rejected more often. Choice B is what a store would like, but an all-must-pass plan cannot do it; a plan that allows one defect can.
Question 9 of 20 · Multiple Choice
In the last minute of an elimination soccer match, a team trailing 1-0 wins a corner kick. An invented model: if the goalkeeper stays back, P(team scores the equalizer) = 0.03 and P(opponent scores) = 0.02; if the goalkeeper joins the attack, these become 0.05 and 0.15. Which analysis is correct?
Answer: A
Only the chance of tying matters to a team that is out if the score stays 1-0, and joining the attack raises it from 0.03 to 0.05. Choices B and C use criteria that ignore the goal: a second goal against changes nothing for the team. Choice D is true about the most likely outcome but misses that one option gives a better chance of the outcome the team needs.
Question 10 of 20 · Multiple Choice
A football team scores a touchdown on the last play of the game and now trails by 1 point. The kick for 1 point succeeds with probability 0.94 and would send the game to overtime, which each team wins half the time. A 2-point try succeeds with probability 0.48 and would win the game. Which choice gives the higher probability of winning?
Answer: D
Kick: P(win) = 0.94(0.5) = 0.47. Two-point try: P(win) = 0.48. The try is slightly better. Choice A forgets that a tie still has to be won in overtime. Choice C doubles 0.48, which gives expected points, not a probability of winning.
Question 11 of 20 · Multiple Choice
An $800 phone has a $120 one-year protection plan. Without the plan, the owner estimates a 15% chance of a $250 screen repair and a separate 5% chance of an $800 replacement (assume both cannot happen). What is the expected cost of going without the plan?
Answer: B
Expected cost = 0.15($250) + 0.05($800) = $37.50 + $40.00 = $77.50, which is $42.50 less than the plan. Choice C adds the costs without weighting them by their probabilities. Choice D is the difference between the plan and the expected cost, not the expected cost itself.
Question 12 of 20 · Multiple Choice
A bank reviews 20,000 card purchases, 100 of which are fraud. Its alert system flags 96 of the fraudulent purchases and 398 of the legitimate ones. What is P(fraud | flagged)?
Answer: C
494 purchases are flagged and 96 of them are fraud: 96/494 ≈ 0.194. Choice A is P(flagged | fraud), the reverse condition. Choice D divides 96 by 398, leaving the fraudulent purchases out of the flagged total. Choice B is the share of legitimate purchases that are flagged.
Question 13 of 20 · Multiple Choice
A condition is serious but easy to treat when it is caught early, and every positive screening result is followed by a second, very accurate test. Which property matters most for the screening test?
Answer: A
A missed case (false negative) goes untreated, while a false positive is caught by the follow-up test, so the screening test should catch as many cases as possible. Choice B would matter more if a positive result led straight to a risky treatment.
Question 14 of 20 · Multiple Choice
To get to an exam that starts in 30 minutes, Route A always takes 25 minutes. Route B takes 20 minutes with probability 0.7 and 40 minutes with probability 0.3. Which route is the better choice?
Answer: C
Route B's expected time is 0.7(20) + 0.3(40) = 26 minutes, and it is late 30% of the time. Route A wins on both criteria. Choice A looks only at the most likely outcome and ignores the 30% chance of being late. Choice B is false: 26 is more than 25.
Question 15 of 20 · Short Answer
A test for a condition has 96% sensitivity and 92% specificity, and 5% of the people tested have the condition. For 10,000 people, find P(condition | positive) and P(no condition | negative). What should happen after a positive result?
500 have the condition: 480 test positive, 20 negative. 9,500 do not: 760 test positive, 8,740 negative. P(condition | positive) = 480/1,240 ≈ 0.387 and P(no condition | negative) = 8,740/8,760 ≈ 0.998. A negative result is very reliable, but most positives are false, so a positive result should lead to a second, confirming test.
Question 16 of 20 · Short Answer
A store accepts a large lot only if all 10 tested items pass. If 3% of the items in the lot are defective, what is the probability that the lot is accepted? Is this a good plan if the store does not want lots with 3% defective?
P(accept) = 0.97¹⁰ ≈ 0.737. The plan accepts such a lot about 74% of the time, so it rarely protects the store from a 3% defect rate. The store would need to test many more items, or accept that the plan targets only much worse lots.
Question 17 of 20 · Short Answer
A hockey game is tied with 2 minutes left. An invented model: with the goalie in net, P(win in regulation) = 0.10, P(lose in regulation) = 0.10 and P(still tied) = 0.80; with the goalie pulled, the three probabilities are 0.18, 0.40 and 0.42. A regulation win is worth 2 standings points and a regulation loss 0. A tie goes to overtime, which each team wins half the time; the overtime winner gets 2 points and the loser 1. Should the team pull the goalie?
A tie is worth 0.5(2) + 0.5(1) = 1.5 expected points. Goalie in: 0.10(2) + 0.80(1.5) + 0.10(0) = 1.40 points. Goalie pulled: 0.18(2) + 0.42(1.5) + 0.40(0) = 0.99 points. Keep the goalie in. When the game is tied, a regulation loss costs points, so the risk of an empty-net goal now matters.
Question 18 of 20 · Short Answer
A school plans an outdoor graduation. Holding it outdoors costs $2,000, plus $6,000 more for a last-minute tent if it rains; the forecast gives a 30% chance of rain. Renting an indoor hall costs $5,000 no matter what. Which plan has the lower expected cost, and at what chance of rain would the two plans cost the same?
Outdoor: $2,000 + 0.30($6,000) = $3,800, less than the $5,000 hall. They cost the same when 2,000 + 6,000p = 5,000, so p = 0.5. The school might still choose the hall if a rainy scramble would ruin the event, a cost the dollar amounts leave out.
Question 19 of 20 · Short Answer
A math contest gives +4 points for a correct answer, -1 for a wrong answer and 0 for a blank. Each question has 5 choices. A student can rule out 2 choices on a question and would guess at random among the rest. Should she guess? What if she cannot rule out any choice?
Guessing among 3 choices: expected points = (1/3)(4) + (2/3)(-1) = 2/3 point > 0, so she should guess. With no choices ruled out: (1/5)(4) + (4/5)(-1) = 0, so guessing and leaving it blank are equal on average.
Question 20 of 20 · Short Answer
A condition affects 0.1% of the population. A screening test has 99% sensitivity and 98% specificity. Find P(condition | positive) for a person from the general population, and explain why doctors do not start treatment on the basis of this screening result alone.
Per 100,000 people: 100 have the condition and 99 test positive; 99,900 do not and 1,998 test positive. P(condition | positive) = 99/2,097 ≈ 0.047. Fewer than 1 in 20 positives are real, so treating every positive would mostly treat healthy people. A positive result should lead to a more specific follow-up test.
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Frequently Asked Questions
10 Questions
What does HSS.MD.B.7 mean?
HSS.MD.B.7 means students can use probability to analyze a decision or a strategy and recommend a choice. The standard's own examples are product testing, medical testing and pulling a hockey goalie, but any situation with uncertain outcomes works. The code stands for High School Statistics and Probability, Using Probability to Make Decisions, cluster B, standard 7.
Is HSS.MD.B.7 taught in Algebra 2 or Statistics?
It is usually taught in a Statistics course or in a probability unit of Algebra II or Precalculus. The "(+)" marks it as one of the additional standards that Common Core lists for students who take advanced courses, so it is not required of every student. It builds directly on the conditional probability standards (HSS.CP) that many schools teach in Geometry or Algebra II.
Why can a positive test result be wrong most of the time even when the test is accurate?
Because when a condition is rare, the few mistakes the test makes on the large healthy group can outnumber the correct results on the small sick group. In Example 1, a test that catches 90% of cases and wrongly flags only 5% of healthy people still gives 980 false positives and 360 true positives in a screening of 20,000 people. That is why a frequency tree with real counts is the clearest way to reason about a positive result.
What is the difference between sensitivity and the chance that a positive result is right?
They are conditional probabilities in opposite directions. Sensitivity is P(positive | condition), which describes the test. The chance that a positive result is right is P(condition | positive), often called the positive predictive value, and it also depends on how common the condition is in the tested group. Confusing the two is a common error that this standard is meant to correct.
Why do hockey coaches pull the goalie, and is it a good decision?
Yes, when the team trails late in the game it is usually a good decision, because it raises the chance of tying. The extra skater also makes an empty-net goal against more likely, but a team that is already losing loses the same standings points by one goal or by two. The right criterion is P(tie), not the expected goal difference. Whether the same reasoning holds in other game situations depends on what each outcome is worth in the standings.
Is expected value always the right way to make a decision?
No: expected value is one criterion, and it fits best when a decision is repeated many times. For a single decision, people also care about the chance of a very bad outcome, about reaching a target (such as staying within a budget) and about costs the model leaves out. Part of HSS.MD.B.7 is choosing a criterion that fits the goal and saying why.
How is product testing a probability decision?
A company cannot test every item, so it tests a sample and decides whether to accept the whole lot. Every plan makes two kinds of error: rejecting a good lot and accepting a bad one. Students compute the probability of each error for different plans, using the multiplication rule for independent items, and weigh them against the cost of testing, especially when testing destroys the item.
What mistakes do students make with HSS.MD.B.7 problems?
Reversing a conditional probability, for example using the sensitivity as the chance that a positive result is right
Dividing by the wrong total in a two-way table, such as the row total when the condition is the column
Choosing the most likely outcome instead of a criterion that fits the goal
Forgetting that a model's probabilities are estimates and that independence is an assumption
Giving a number with no recommendation
Do students need Bayes' theorem for this standard?
No: a frequency tree or a two-way table with a round population gives the same answer, and it shows where the answer comes from. Students who want a formula can write P(condition | positive) = (sensitivity × prevalence) / (sensitivity × prevalence + (1 - specificity)(1 - prevalence)), which is Bayes' theorem for this case. Many teachers introduce the formula only after students have built several trees.
How does HSS.MD.B.7 connect to other standards and tests?
It applies conditional probability and two-way tables (HSS.CP.A.4, HSS.CP.B.6) and works alongside the expected-value standards HSS.MD.B.5 and fair decisions in HSS.MD.B.6. It leads into statistical inference, where simulation is used to judge a model (HSS.IC.A.2). The digital SAT's Problem-Solving and Data Analysis domain includes conditional probability questions from two-way tables, which use the same skill as the medical testing problems here.
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Related Standards
6 standards
These standards connect to HSS.MD.B.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSS.CP.A.4Prerequisite
Build and read two-way frequency tables and use them to find conditional probabilities
