L-PR-2 · GUIDED INQUIRY Introduction to Probability and Statistics

L-PR-2 · Guided inquiry

What does a positive test really tell us?

Conditional Probability

How does “given” change both the question and the denominator?

Anonymous check

After a positive result, what is the probability the person has the condition?

In a hypothetical population, a condition affects 1% of people. The test detects 95% of people with the condition (sensitivity) and correctly gives a negative result to 95% without it (specificity).

  1. 95%

  2. 50%

  3. About 16.1%

  4. The information is insufficient

Inquiry map

What must we learn to answer “given” questions?

Eight concepts in the lesson’s approved order.

C1 → C2 Restrict, then renormalize

Interpret conditioning and choose the correct denominator.

C3 → C4 Build joint paths

Use the General Multiplication Rule and probability trees.

C5 → C7 Test relationships

Verify independence and separate it from mutual exclusivity.

C8 Protect the direction

Do not confuse a conditional with its reverse.

The event after “given” defines the new universe.

Anonymous check

What is P(group ∩ passed)?

Among 80 students, 36 studied in a group and passed, 45 studied in a group, and 60 passed.

  1. 36/80

  2. 36/45

  3. 36/60

  4. 45/80

Investigation 1 · Restrict the sample space

C1 · What changes when the universe shrinks?

A bag has \(20\) marbles: \(8\) red, \(6\) blue, and \(6\) green.

Investigate: If the marble is known not to be green, what remains possible?

  1. Original universe
    \(20\)

    Every marble is initially possible.

  2. Condition
    \(\text{not green}\)

    Remove the \(6\) green marbles.

  3. Restricted universe
    \(14\)

    The \(8\) red marbles remain among \(14\).

\(P(\text{red}\mid\text{not green})=8/14=4/7\approx 0.57\).

Conditional probability · Concept

C1 · Conditioning is restrict and renormalize.

When \(P(B)>0\), conditioning on \(B\) discards outcomes outside \(B\).

Investigate: What must become the new whole?

  1. Restrict
    \(B\)

    Keep only outcomes in \(B\).

  2. Favourable outcomes
    \(A\cap B\)

    Inside the new universe, these outcomes also satisfy \(A\).

  3. Renormalize
    \(P(B\mid B)=1\)

    Probabilities within the restricted universe must total one.

\(P(A\mid B)\) means the probability of \(A\) after learning that \(B\) occurred.

Anonymous check

Which denominator belongs in P(passed | group)?

In a survey, 36 group-studiers passed, 45 studied in a group, 60 passed, and 80 were surveyed.

  1. 80

  2. 45

  3. 36

  4. 60

Conditional probability · Formula and table method

C2 · The formula records the restricted universe.

For \(P(B)>0\), the conditioning event supplies the denominator.

  1. Formula
    \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)

    Joint probability over the probability of the known event.

  2. Count form
    \(\frac{\text{count in }A\cap B}{\text{marginal total for }B}\)

    Use the row or column total for the event after “given.”

  3. Study table
    \(P(\text{passed}\mid\text{group})=36/45=0.80\)

    The group row becomes the whole.

Choose the denominator from the language before calculating.

Conditioning direction

C2 · Same intersection. Different questions.

Both directions use the \(36\) students in the intersection.

Investigate: Why do the answers differ?

Passed given group \(36/45=0.80\)

The universe is the \(45\) group-studiers.

Group given passed \(36/60=0.60\)

The universe is the \(60\) students who passed.

\(P(A\mid B)\) and \(P(B\mid A)\) share a numerator but generally not a denominator.

Prediction · Drawing without replacement

C3 · How do we find the probability of a path?

A bag contains \(5\) red and \(3\) blue marbles. Two are drawn without replacement.

Investigate: What changes after the first marble is red?

  1. First branch
    \(P(R_1)=5/8\)

    Five of the eight original marbles are red.

  2. Second branch
    \(P(B_2\mid R_1)=3/7\)

    Seven marbles remain, including all three blue.

  3. Complete path
    \(P(R_1\cap B_2)=(5/8)(3/7)=15/56\)

    Multiply the unconditional and conditional branches.

\(15/56\approx 0.268\). Without replacement, the second probability is conditional.

Derive, then use

C3 · The General Multiplication Rule works for dependent events.

Start from the conditional probability formula with \(P(B)>0\).

Investigate: Which operation isolates the joint probability?

  1. Start
    \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)

    The joint probability is divided by the conditioning probability.

  2. Multiply
    \(P(B)P(A\mid B)=P(A\cap B)\)

    Multiply both sides by \(P(B)\).

  3. Symmetric form
    \(P(A\cap B)=P(A)P(B\mid A)\)

    Use this direction when P(A) > 0 and P(B | A) is supplied.

Use the simplified product only after independence is established.

Tree diagrams · Labeling convention

C4 · What belongs on each part of a probability tree?

A tree makes the conditioning history visible.

Investigate: Which quantities belong on branches—and which belong at path ends?

Swipe across the tree to see the joint outcomes.

Two-stage probability tree structure A two-stage tree. Branches show conditional history and leaves show joint outcomes. Start P(L₁) P(L₂) P(D | L₁) P(good | L₁) P(D | L₂) P(good | L₂) P(L₁ ∩ D) P(L₁ ∩ good) P(L₂ ∩ D) P(L₂ ∩ good)

Branches from one node sum to one; paths to the same event add.

Worked tree · Two assembly lines

C4 · Multiply along paths. Add across paths.

Line \(1\) makes \(60\%\) of parts with a \(2\%\) defect rate; Line \(2\) makes \(40\%\) with a \(5\%\) defect rate.

Investigate: Which paths end in a defective part?

Swipe across the tree to see the joint outcomes.

Four-path factory probability tree A two-stage tree. Branches show conditional history and leaves show joint outcomes. Start Line 1 · 0.60 Line 2 · 0.40 D · 0.02 Good · 0.98 D · 0.05 Good · 0.95 0.60 × 0.02 = 0.012 0.60 × 0.98 = 0.588 0.40 × 0.05 = 0.020 0.40 × 0.95 = 0.380

Defective parts: 0.012 + 0.020 = 0.032

All four paths: 0.012 + 0.588 + 0.020 + 0.380 = 1.

Anonymous check

What is the probability that a random parcel is flagged?

Route A carries 70% of parcels and flags 3% of its parcels. Route B carries 30% and flags 8% of its parcels.

  1. 0.045

  2. 0.065

  3. 0.000504

  4. 0.11

Independent events · Definition

C5 · When does conditioning provide no new information?

Independence is a probability relationship—not a judgment that events “sound unrelated.”

Investigate: What should remain unchanged after learning that the other event occurred?

  1. Conditional form
    \(P(A\mid B)=P(A)\)

    Valid when \(P(B)>0\).

  2. Symmetry
    \(P(B\mid A)=P(B)\)

    Valid when \(P(A)>0\).

  3. Product form
    \(P(A\cap B)=P(A)P(B)\)

    An equivalent definition and test.

If the conditional and marginal probabilities differ, the events are dependent.

Anonymous check

What relationship does the data support?

Among 300 residents, 180 have a gym membership, 120 report healthy eating, and 72 have both.

  1. Mutually exclusive

  2. Independent

  3. Both mutually exclusive and independent

  4. Neither

Two equivalent independence tests

C6 · Verify before using the simplified rule.

The city survey supplies all values needed for both tests.

Investigate: Do the product and conditional checks agree?

  1. Observed joint
    \(P(G\cap H)=72/300=0.24\)

    Read the intersection from the data.

  2. Product test
    \(P(G)P(H)=0.60(0.40)=0.24\)

    Observed and expected intersections agree.

  3. Conditional test
    \(P(G\mid H)=72/120=0.60=P(G)\)

    Conditioning leaves the probability unchanged.

The two tests agree: the events are independent in this sample, without implying causation.

Prediction · Mutually exclusive versus independent

C7 · Can “cannot happen together” mean “independent”?

Suppose \(P(A)>0\), \(P(B)>0\), and \(A\cap B=\varnothing\).

Investigate: After learning \(B\) occurred, what happens to the probability of \(A\)?

Before conditioning \(P(A)>0\)

Event \(A\) is possible.

After conditioning on B \(P(A\mid B)=0\)

Event \(B\) rules \(A\) out.

The probability changes, so positive-probability mutually exclusive events are dependent.

Proof and edge case

C7 · Non-overlap is extreme information.

Use the conditional definition instead of relying on labels.

  1. Mutually exclusive
    \(P(A\cap B)=0\)

    The events cannot co-occur.

  2. Condition on B
    \(P(A\mid B)=\frac{0}{P(B)}=0\)

    This is defined when \(P(B)>0\).

  3. Compare
    \(0\ne P(A)\)

    Because \(P(A)>0\), independence fails.

Only a zero-probability event edge case can be both mutually exclusive and independent.

Impossible-event edge case

C7 · When can both labels be true?

Let A be impossible and let P(B) = 0.50.

Investigate: Test each relationship separately.

  1. No overlap
    P(A ∩ B) = 0

    A cannot happen with B: mutually exclusive.

  2. Product test
    P(A)P(B) = 0 × 0.50 = 0

    The joint equals the product: independent.

  3. Conditioning limit
    P(B | A) is undefined

    Use the product test when the conditioning event is impossible.

This edge case is both; the positive-probability non-overlap rule remains dependent.

Anonymous check

How should the pair be classified?

P(A) = 0.30, P(B) = 0.50, and P(A ∩ B) = 0.15.

  1. Mutually exclusive

  2. Independent

  3. Both

  4. Neither

Asymmetry of conditioning

C8 · Which group does each conditional describe?

The order around the conditioning bar determines the question.

Investigate: Whose universe is in the denominator?

Sensitivity \(P(\text{positive}\mid\text{condition})=0.95\)

Among people who have the condition, how many test positive?

Positive predictive value \(P(\text{condition}\mid\text{positive})\)

Among everyone who tested positive, how many have the condition?

Reversing the conditioning direction changes both the denominator and the meaning.

Rare-condition calculation

C8 · Why can an accurate test surprise us?

Hypothetical aggregate illustration, not medical advice: use prevalence \(0.01\), sensitivity \(0.95\), and false-positive rate \(0.05\).

Investigate: Which joint paths end in a positive test?

True-positive path \(0.01(0.95)=0.0095\)

Condition and positive.

False-positive path \(0.99(0.05)=0.0495\)

No condition and positive.

All positives \(P(+)=0.0095+0.0495=0.059\)

Add both positive-ending paths.

Reverse conditional \(P(\text{condition}\mid +)=\frac{19}{118}\approx 0.161\)

The true-positive path is divided by the probability of all positive results.

High sensitivity does not erase a low base rate.

Anonymous check

What does the statement actually tell us?

In a hypothetical screening example, sensitivity is 95%: the test is positive for 95% of people who have the condition.

  1. Among people who test positive, 95% have the condition

  2. Among people with the condition, 95% test positive

  3. Both directions are always equal

  4. It is the overall accuracy rate

Remote work and tenure

Worked comparison · One table, two directions.

Among \(200\) employees, \(54\) both prefer remote work and have tenure above \(5\) years; \(90\) have long tenure and \(120\) prefer remote work.

Investigate: Which marginal total belongs to each question?

  1. Remote given long tenure
    \(54/90=0.60\)

    The long-tenure row is the restricted universe.

  2. Long tenure given remote
    \(54/120=0.45\)

    The remote-work column is the restricted universe.

  3. Comparison
    \(0.60\ne 0.45\)

    The numerator stays fixed while the denominator changes.

Translate the conditioning direction before reading the table.

Factory defects

Worked application · Reverse a tree result.

The factory tree gave \(P(D)=0.032\) and \(P(L_1\cap D)=0.012\).

Investigate: Given a part is defective, what is \(P(L_1\mid D)\)?

  1. Restrict
    \(D\)

    Only defective parts remain in the denominator.

  2. Intersection
    \(L_1\cap D\)

    The numerator is the Line 1 defective path.

  3. Compute
    \(P(L_1\mid D)=0.012/0.032=3/8=0.375\)

    Line 1 makes more parts but a smaller share of the defective group.

Tree paths create the joint probabilities needed for a reverse conditional.

Conditioning-direction diagnosis

Find the error · What does the 80% condition on?

“Among people who exercised regularly, \(80\%\) reported good sleep.”

Investigate: Is the stated value enough to compute the reverse conditional?

  1. Correct translation
    \(P(\text{sleep}\mid\text{exercise})=0.80\)

    The reference group is regular exercisers.

  2. Student’s reversal
    \(P(\text{exercise}\mid\text{sleep})=0.80\)

    This silently changes the denominator.

  3. Missing information
    \(P(\text{exercise}),\;P(\text{sleep})\)

    Base rates are needed to compute the reverse direction.

A conditional probability cannot be reversed by swapping its labels.

Guided recap · Set up → calculate → classify

Which evidence supports each step?

Revisit the worked examples before the fresh exit checks.

Investigate: For each item, name the evidence that fixes the denominator or relationship.

  1. 1 · Conditional
    \(P(\text{remote}\mid\text{long tenure})\)

    Set it up from \(54\) in the intersection and \(90\) long-tenure employees.

  2. 2 · Tree and reverse
    \(P(D),\;P(L_1\mid D)\)

    Use defective paths \(0.012\) and \(0.020\).

  3. 3 · Classify
    \(P(A)=0.35,\;P(B)=0.40,\;P(A\cap B)=0.20\)

    Choose mutually exclusive, independent, both, or neither—and justify.

Targets: \(0.60\); \(0.032\) and \(0.375\); neither, because \(0.20\ne 0.35(0.40)=0.14\) and \(0.20>0\).

Anonymous check

What is P(brought notes | completed task)?

At a workshop, 120 people attended. Fifty brought notes; 30 of those completed the task. Forty-two people completed the task in total.

  1. 30/42

  2. 30/50

  3. 42/120

  4. 50/120

Anonymous check

Are the events mutually exclusive, independent, both, or neither?

For events A and B, P(A) = 0.30, P(B) = 0.40, and P(A ∩ B) = 0.05.

  1. Mutually exclusive

  2. Independent

  3. Both

  4. Neither

Closing synthesis

“Given” is an instruction to rebuild the universe.

Carry these four checks into every conditional-probability problem.

Direction What is known?

The event after “given” sets the denominator.

Joint path What must happen together?

Multiply unconditional and conditional branches.

Relationship Did the probability change?

Test independence; do not assume it from context.

Classification Together—or unchanged?

Mutual exclusivity and independence require different evidence.

A defensible answer names its restricted universe, calculation, and relationship evidence.

Anonymous exit reflection · confidence, not mastery

How confident are you with each main idea?

Use your exit responses as evidence. Rate each statement from \(0\) to \(5\), then submit once.

C1–C2 + C8

I can set up and calculate a conditional probability in the correct direction.

C3

I can calculate a joint probability with the General Multiplication Rule.

C4

I can build and use a probability tree for a multi-stage event.

C5–C7

I can classify events using intersection and independence evidence.

\(0\) Not yet \(1\) Very unsure \(2\) Beginning \(3\) Mostly understand \(4\) Confident \(5\) I could explain it