Build independence from a grid whose columns are one event and whose rows are the other, derive all three tests from it, separate mutually exclusive events from independent ones, and work draws with and without replacement.
Subject: Statistics · 62 slides · applied lesson
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Title
Statistics · §3.2
Three tests for independence, and why exclusive is its opposite
Objectives
Concept
U, V: events · U′: not U
\[ P(U) = \tfrac{\text{count } U}{\text{count } S} \]
Count
Why: S holds every equally likely outcome
\[ P(U') = 1 - P(U) \]
Complement
Why: U and U′ together fill S
\[ P(U \mid V) = \tfrac{P(U \text{ AND } V)}{P(V)} \]
News about V
Why: V becomes the new sample space
\[ \tfrac{u}{v} = \tfrac{u \div w}{v \div w} \]
Scale
Why: same share, fewer larger parts
\[ \tfrac{u}{v} \cdot \tfrac{w}{z} = \tfrac{uw}{vz} \]
Multiply shares
Why: tops and bottoms multiply apart
\[ u \text{ cols} \times v \text{ rows} = uv \text{ cells} \]
Rectangle count
Why: each column meets each row once
\[ (u \times v) \div u = v \ \ (u \ne 0) \]
Cancel
Why: dividing undoes the multiplying
\[ 2 \div 3 \approx 0.67 \]
Round
Why: ≈ marks a rounded decimal
Prediction
Figure (svg): A grid of 200 cells in 20 columns and 10 rows, one student each; the first 15 columns filled blue for the 150 who take English; the first 3 rows outlined in heavy ink for the 60 who take speech; the 45 cells in both, 15 wide and 3 tall, filled orange
Predict first
Example 3.10: of 200 students, 150 take English, 60 take speech, 45 take both.
The classmate says the two classes are so different that nobody takes both. What does the grid settle?
Correct: That 45 students do take both
Why: The orange block is 45 of the 200 cells, and a cell is one student, so the claim that nobody takes both is already dead. Whether hearing that a student takes speech moves the chance they take English is a second, different question, and this deck builds the test that answers it.
Worked example
Figure (svg): A grid of 200 cells in 20 columns and 10 rows, one student each; the first 15 columns filled blue for the 150 who take English; the first 3 rows outlined in heavy ink for the 60 who take speech; the 45 cells in both, 15 wide and 3 tall, filled orange
\[ 15 \times 3 = \textcolor{#b54708}{45} \]
Count the orange cells
Why: columns times rows fills a rectangle
\[ P(C \text{ AND } D) = \frac{\textcolor{#b54708}{45}}{200} \]
Write it as a chance
Why: every student is equally likely
\[ \frac{\textcolor{#b54708}{45}}{200} = \textcolor{#b54708}{0.225} \]
Divide 45 by 200
Why: a decimal to set against 0
\[ \textcolor{#b54708}{0.225} \ne 0 \]
Compare with zero
Why: kills the nobody-takes-both claim
\[ 200 \times \textcolor{#b54708}{0.225} = \textcolor{#b54708}{45} \]
Check: rebuild the count
Why: multiplying undoes the division
Section
Idea 1 of 4
Concept
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
A survey of 100 students: 60 take math, 50 take science, 30 take both.
Discussion prompt
What two numbers would you set side by side to answer the advisor?
Answer:
Math's share among the science students, against math's share among all 100.
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
\[ \textcolor{#b54708}{30} < \textcolor{#1f5fbf}{60} \]
Set the two counts together
Why: the comparison an advisor reaches for
\[ \textcolor{#b54708}{30} \text{ out of } 50 \]
Name the block's group
Why: the science students, not everyone
\[ \textcolor{#1f5fbf}{60} \text{ out of } 100 \]
Name the columns' group
Why: a different sized group entirely
\[ 50 \ne 100 \]
Compare the two groups
Why: fails: counts from unequal groups
Counts need their groups before they can be compared.
\[ \textcolor{#b54708}{30} + 20 = 50 \]
Check: fill the science rows
Why: 20 science students skip math
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
\[ \frac{\textcolor{#b54708}{30}}{50} = \textcolor{#b54708}{0.6} \]
Divide the block by 50
Why: math's share inside the science group
\[ \frac{\textcolor{#1f5fbf}{60}}{100} = \textcolor{#1f5fbf}{0.6} \]
Divide the columns by 100
Why: math's share of everybody
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange; every cell outside the 5 science rows faded and dashed
\[ \textcolor{#b54708}{0.6} = \textcolor{#1f5fbf}{0.6} \]
Set the two shares together
Why: the news moved nothing at all
Blue runs in whole columns, so it cuts every row alike.
\[ 0.6 \times 50 = \textcolor{#b54708}{30},\ \ 0.6 \times 100 = \textcolor{#1f5fbf}{60} \]
Check: rebuild both counts
Why: one share fits both groups
Worked example
Figure (svg): Two bars of equal length, one for the 50 science students and one for the 50 others; each has 30 of its 50 filled blue for the math takers
\[ P(A) = \textcolor{#1f5fbf}{0.6} \]
Name math's own chance
Why: the baseline the test compares against
\[ P(A \mid B) = \textcolor{#b54708}{0.6} \]
Name the chance given B
Why: what hearing B does to that baseline
\[ P(A \mid B) = P(A) \]
Match the two chances
Why: hearing B was worth nothing here
Figure (svg): Two bars of equal length, one for the 50 science students and one for the 50 others; each has 30 of its 50 filled blue for the math takers, with a grey dashed line at the 0.6 share that cuts both bars at the same place
\[ 0.6 \times 50 = 30 \]
Check either bar against 0.6
Why: each group of 50 holds 30
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science
\( {P(A \mid B) = P(A)} \)
\[ P(A \mid B) = \textcolor{#b54708}{P(A \text{ AND } B)} \div P(B) \]
Write the conditional rule
Why: the only link to the overlap
\[ \textcolor{#1f5fbf}{P(A)} = \textcolor{#b54708}{P(A \text{ AND } B)} \div P(B) \]
Substitute P(A) on the left
Why: independence says the two are equal
\[ \textcolor{#1f5fbf}{P(A)} \cdot P(B) = \textcolor{#b54708}{P(A \text{ AND } B)} \]
Multiply both sides by P(B)
Why: clears the fraction off the right
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
\[ \textcolor{#1f5fbf}{6} \times 5 = \textcolor{#b54708}{30} \text{ cells} \]
Count the orange rectangle
Why: columns times rows fills it
Figure (svg): The 100-cell grid with the 30-cell orange block, a blue bar 6 columns long above it marked 6 wide and an ink bar 5 rows long beside it marked 5 tall
\[ \frac{\textcolor{#b54708}{30}}{100} = \frac{\textcolor{#1f5fbf}{6} \times 5}{10 \times 10} \]
Write both counts as products
Why: a rectangle's area splits in two
\[ \frac{\textcolor{#1f5fbf}{6} \times 5}{10 \times 10} = \frac{\textcolor{#1f5fbf}{6}}{10} \cdot \frac{5}{10} \]
Split width from height
Why: tops and bottoms multiply apart
Width share × height share = the block's share of the grid.
\[ \textcolor{#1f5fbf}{0.6} \times 0.5 = \textcolor{#b54708}{0.3} \]
Check against the counted block
Why: 30 of 100 cells is 0.3
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange
\( {P(A \text{ AND } B) = P(A) \cdot P(B)} \)
\[ P(B \mid A) = \textcolor{#b54708}{P(A \text{ AND } B)} \div \textcolor{#1f5fbf}{P(A)} \]
Write B's conditional rule
Why: the two events swap roles
\[ P(B \mid A) = \frac{\textcolor{#1f5fbf}{P(A)} \cdot P(B)}{\textcolor{#1f5fbf}{P(A)}} \]
Substitute the product
Why: sets up the cancelling step
\[ P(B \mid A) = P(B) \]
Cancel the P(A) factors
Why: dividing undoes the multiplying
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange; every cell outside the 6 math columns faded and dashed
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange; every cell outside the 5 science rows faded and dashed
\[ \begin{aligned} P(A \mid B) &= P(A) \\ P(B \mid A) &= P(B) \\ P(A \text{ AND } B) &= P(A)P(B) \end{aligned} \]
\[ \textcolor{#1f5fbf}{0.6} = \textcolor{#1f5fbf}{0.6},\ \ 0.5 = 0.5 \]
Read the two news tests
Why: each chance survived its own news
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange; every cell outside the 6 math columns faded and dashed
\[ \textcolor{#1f5fbf}{0.6} \times 0.5 = \textcolor{#b54708}{0.3} \]
Read the third test
Why: the overlap is exactly the product
\[ \textcolor{#b54708}{0.3} \times 100 = \textcolor{#b54708}{30} \text{ cells} \]
Check: turn the product into cells
Why: the orange block holds exactly 30
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 who take math; the first 5 rows outlined in heavy ink for the 50 who take science; the 30 cells in both, 6 wide and 5 tall, filled orange; every cell outside the 5 science rows faded and dashed
G: math class · H: science class
Given: P(G) = 0.6, P(H) = 0.5, P(G AND H) = 0.3
\[ P(G \mid H) = \frac{\textcolor{#b54708}{0.3}}{0.5} \]
Put the overlap over P(H)
Why: the science group is the total
\[ \frac{\textcolor{#b54708}{0.3}}{0.5} = \textcolor{#1f5fbf}{0.6} \]
Divide 0.3 by 0.5
Why: a number to set beside P(G)
\[ \textcolor{#1f5fbf}{0.6} = P(G) \]
Compare with P(G)
Why: independence needs these two equal
\[ P(G)P(H) = \textcolor{#1f5fbf}{0.6} \times 0.5 = \textcolor{#b54708}{0.3} \]
Check with the product test
Why: a route that never divides
Trap
Figure (svg): A bar for the 60 math students with 30 of them filled blue for science
\[ P(H \mid G) = \textcolor{#b54708}{0.3} \div \textcolor{#1f5fbf}{0.6} = 0.5 \]
Divide by P(G)
Why: math is now the whole group
\[ 0.5 \ne \textcolor{#1f5fbf}{0.6} \]
Compare with P(G|H)
Why: fails: they need not match
Figure (svg): A bar for the 50 science students with 30 of them filled blue for math
\[ P(G \mid H) = \textcolor{#1f5fbf}{0.6} = P(G) \]
Match math with its own chance
Why: the comparison the test wants
\[ P(H \mid G) = 0.5 = P(H) \]
Check the other pair too
Why: it also keeps its own chance
Prediction
Figure (svg): A chance scale from 0 to 1 with marks at P(walk) = 0.3 and P(walk | car) = 0.3
Predict first
In a campus study, 0.3 of students walk to campus, and among the car owners the walking share is also 0.3.
What follows from those two figures alone?
Correct: Walking and owning a car are independent
Why: The news that a student owns a car left the walking chance at 0.3, which is the first test passing. The other three options each describe a change the figures do not show: a share of 0.3 among car owners is neither zero nor below the overall 0.3.
Worked example
Figure (svg): A chance scale from 0 to 1 with marks at P(walk) = 0.3 and P(walk | car) = 0.3
\[ P(\text{walk}) = \textcolor{#1f5fbf}{0.3},\ \ P(\text{walk} \mid \text{car}) = \textcolor{#b54708}{0.3} \]
Write the two given shares
Why: the study reports both directly
\[ P(\text{walk} \mid \text{car}) = P(\text{walk}) \]
Match them
Why: independence asks exactly this
\[ \textcolor{#b54708}{0.3} \ne 0,\ \ \textcolor{#b54708}{0.3} \not< \textcolor{#1f5fbf}{0.3} \]
Compare with 0 and 0.3
Why: rules out two rival options
\[ P(\text{walk AND car}) = P(\text{car}) \times \textcolor{#b54708}{0.3} \]
Check: the overlap is positive
Why: any car owners make it nonzero
Faded example
Figure (svg): A chance scale from 0 to 1 with marks at P(A)P(B) and P(A AND B)
Try It 3.8: P(A) = 0.4, P(B) = 0.2, P(A AND B) = 0.08.
A: learning Spanish · B: learning German
Fill in the blanks
P(A)P(B) = 0.08; P(A AND B) = 0.08, so A and B are independent
Why: 0.4 × 0.2 = 0.08, which is exactly the given P(A AND B), so the product test passes and the two events are independent: hearing that a student learns German leaves the chance of Spanish at 0.4.
Worked example
Figure (svg): A chance scale from 0 to 1 with marks at P(A)P(B) and P(A AND B)
\[ P(A)P(B) = 0.4 \times 0.2 \]
Write the product test
Why: the given facts allow this one
\[ 0.4 \times 0.2 = \textcolor{#1f5fbf}{0.08} \]
Multiply the two chances
Why: the overlap independence predicts
\[ \textcolor{#1f5fbf}{0.08} = \textcolor{#b54708}{0.08} \]
Compare with the given overlap
Why: the test asks for exactly this match
So A and B are independent.
\[ P(A \mid B) = \frac{\textcolor{#b54708}{0.08}}{0.2} = 0.4 \]
Check with the news test
Why: Spanish keeps its own chance
Section
Idea 2 of 4
Concept
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 women; the long-haired students outlined in heavy ink at the top of each column, 8, 8, 8, 7, 7, 7 deep in the women's columns and 2, 1, 1, 1 deep in the men's
Example 3.12: 45 of the 100 are women with long hair.
Discussion prompt
The ink outline runs far deeper in the women's columns. What does that already suggest?
Answer:
Hearing that a student is a woman changes the chance of long hair.
Worked example
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 women; the long-haired students outlined in heavy ink at the top of each column, 8, 8, 8, 7, 7, 7 deep in the women's columns and 2, 1, 1, 1 deep in the men's
\[ \textcolor{#1f5fbf}{60} \div 100 = \textcolor{#1f5fbf}{0.6},\ \ 50 \div 100 = 0.5 \]
Divide each group by 100
Why: width share and height share
\[ \textcolor{#1f5fbf}{0.6} \times 0.5 = \textcolor{#b54708}{0.3} \]
Multiply the two shares
Why: the overlap a rectangle would have
Figure (svg): A 10 by 10 grid of 100 cells, one student each; the first 6 columns filled blue for the 60 women; the long-haired students outlined in heavy ink at the top of each column, 8, 8, 8, 7, 7, 7 deep in the women's columns and 2, 1, 1, 1 deep in the men's; a dashed orange rectangle 6 wide and 5 tall marking the 30 cells independence would need
\[ \textcolor{#b54708}{0.3} \times 100 = \textcolor{#b54708}{30} \]
Turn the share into a count
Why: a count the grid can check
\[ \textcolor{#b54708}{30} \ne 45 \]
Compare with the counted 45
Why: fails: 15 more than predicted
\[ 45 - \textcolor{#b54708}{30} = 15 \]
Check: measure the miss
Why: 15 cells stick out
Worked example
Figure (svg): Two bars of equal length, one for the 60 women with 45 filled blue and one for the 40 men with 5 filled blue
\[ \frac{45}{\textcolor{#1f5fbf}{60}} = 0.75 \]
Divide 45 by the women
Why: long hair's share inside the women
\[ 50 - 45 = 5 \]
Subtract to find their count
Why: the other long-haired students
\[ 100 - \textcolor{#1f5fbf}{60} = 40 \]
Subtract to find the men
Why: the group those 5 belong to
\[ \frac{5}{40} = 0.125 \]
Divide 5 by the men
Why: the same share, other group
Figure (svg): Two bars of equal length, one for the 60 women with 45 filled blue and one for the 40 men with 5 filled blue, and a grey dashed line at the 0.5 share, which cuts the two bars in different places
\[ 0.75 \ne 0.125 \]
Set the two shares together
Why: the news moves the chance far
\[ 0.75 \times \textcolor{#1f5fbf}{60} = 45 \]
Check: undo the women's division
Why: returns the count the survey gave
Worked example
Figure (svg): Two bars of equal length, one for the 60 women with 45 filled blue and one for the 40 men with 5 filled blue, and a grey dashed line at the 0.5 share, which cuts the two bars in different places
\( {P(L \mid W) = 0.75}\;\;\Rightarrow\;\;\allowbreak {P(L) = 0.5} \)
\[ \textcolor{#1f5fbf}{0.75} \ne 0.5 \]
Compare the news test's sides
Why: hearing 'woman' raised the chance
\[ P(W)P(L) = 0.6 \times 0.5 = \textcolor{#b54708}{0.3} \]
Run the product test
Why: a product to set against 0.45
\[ \textcolor{#b54708}{0.3} \ne 0.45 \]
Compare with the count
Why: it rejects independence too
\[ P(W \mid L) = \frac{0.45}{0.5} = 0.9 \ne 0.6 \]
Check with the reversed news
Why: hearing L must move W as well
Trap
Figure (svg): A bar of 100 students headed the rule says 30, with 30 of the 100 filled blue
\[ \textcolor{#1f5fbf}{0.6} \times 0.5 = \textcolor{#b54708}{0.3} \]
Apply the product rule
Why: as though it always held
\[ \textcolor{#b54708}{30} \ne 45 \]
Compare with the survey
Why: fails: 45 were counted
Figure (svg): A bar of 100 students headed the survey found 45, with 45 of the 100 filled blue
\[ P(W \text{ AND } L) = 0.45 \ne \textcolor{#b54708}{0.3} \]
Take the counted share
Why: the rule wants independence first
\[ 0.45 \div 0.6 = 0.75 \ne 0.5 \]
Check the news test
Why: a second route to the verdict
Prediction
Figure (svg): Two bars of equal length, one for the 60 women with 45 filled blue and one for the 40 men with 5 filled blue, and a grey dashed line at the 0.5 share, which cuts the two bars in different places
Predict first
Keep 60 women, 40 men and 50 long-haired students among the 100.
How many of the 40 men would need long hair to make the two events independent?
Correct: 20
Why: Independence needs long hair to take the same share of each group, and that share has to be the class-wide 50 of 100, or 0.5. Half of 40 is 20 men, half of 60 is 30 women, and 20 + 30 rebuilds the 50 long-haired students.
Worked example
Figure (svg): Two bars of equal length, one for the 60 women with 30 filled blue and one for the 40 men with 20 filled blue
\[ \frac{50}{100} = 0.5 \]
Divide the long-haired by 100
Why: the share each group must match
\[ 0.5 \times 40 = \textcolor{#1f5fbf}{20} \]
Take that share of the men
Why: the count the question asks for
\[ 0.5 \times 60 = \textcolor{#1f5fbf}{30} \]
Take that share of the women
Why: both groups must match the class
Figure (svg): Two bars of equal length, one for the 60 women with 30 filled blue and one for the 40 men with 20 filled blue, and a grey dashed line at the 0.5 share cutting both bars at the same place
\[ \textcolor{#1f5fbf}{20} + \textcolor{#1f5fbf}{30} = 50 \]
Add the two counts
Why: they must rebuild the 50 given
\[ P(W)P(L) = 0.6 \times 0.5 = \frac{\textcolor{#1f5fbf}{30}}{100} \]
Check with the product test
Why: 30 long-haired women, as built
Faded example
Figure (svg): A chance scale from 0 to 1, headed "B: book, D: DVD", marked at P(B) = 0.40 and P(B | D) ≈ 0.67
Try It 3.10: P(B) = 0.40, P(D) = 0.30, P(B AND D) = 0.20.
B: checks out a book · D: checks out a DVD
Fill in the blanks
P(B | D) = 0.20 ÷ 0.30 ≈ 0.67; P(D | B) = 0.20 ÷ 0.40 = 0.5; independent? no
Why: 0.20 ÷ 0.30 ≈ 0.667 and 0.20 ÷ 0.40 = 0.5. Neither matches its own event's chance: 0.667 is not 0.40 and 0.5 is not 0.30, so the events are dependent. The product test says the same, since 0.40 × 0.30 = 0.12, not 0.20.
Worked example
Figure (svg): A chance scale from 0 to 1, headed "B: book, D: DVD", marked at P(B) = 0.40 and P(B | D) ≈ 0.67
\[ \begin{aligned} P(B \mid D) &= 0.20 \div 0.30 \\ P(D \mid B) &= 0.20 \div 0.40 \end{aligned} \]
Put the overlap over each chance
Why: each borrower group becomes a total
\[ \begin{aligned} 0.20 \div 0.30 &\approx \textcolor{#b54708}{0.67} \\ 0.20 \div 0.40 &= 0.5 \end{aligned} \]
Divide both
Why: decimals for the two comparisons
\[ \textcolor{#b54708}{0.67} \ne \textcolor{#1f5fbf}{0.40},\ \ 0.5 \ne 0.30 \]
Compare each with its own
Why: both news tests reject it
\[ \textcolor{#1f5fbf}{0.40} \times 0.30 = 0.12 \ne 0.20 \]
Check with the product test
Why: the product against the same overlap
Section
Idea 3 of 4
Concept
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both
Try It 3.12: P(I) = 0.44 and P(F) = 0.56, and Mark takes one route.
I: takes the Interstate · F: takes Fifth Street
Discussion prompt
What is the chance that a single morning is an Interstate morning and a Fifth Street morning?
Answer:
No morning can be both, so no cell in the strip carries both routes.
Worked example
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both
\[ 0.44 + 0.56 = 1 \]
Add Mark's two chances
Why: a sum of 1 suggests no room
Made up: 44 of 100 lunches take soup, 56 take salad, 20 take both.
\[ 0.44 + 0.56 = 1 \]
Add the canteen's chances
Why: the same sum from different data
Figure (svg): A strip of 100 made-up lunches: 44 cells filled blue for soup and 56 outlined in ink for salad, with 20 cells in both filled orange
\[ P(\text{soup AND salad}) = \textcolor{#b54708}{0.20} \]
Read the counted overlap
Why: yet one sum gave two answers
\[ 44 + 56 - \textcolor{#b54708}{20} = 80 \]
Check: remove the double count
Why: 80 lunches take at least one dish
Worked example
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both
\[ \textcolor{#1f5fbf}{44} + 56 = 100 \]
Add the two blocks
Why: they already use every morning
\[ 100 - 100 = \textcolor{#b54708}{0} \]
Subtract from 100 cells
Why: no cell is left for both
\[ P(I \text{ AND } F) = \frac{\textcolor{#b54708}{0}}{100} = \textcolor{#b54708}{0} \]
Write the overlap as a chance
Why: zero outcomes gives chance zero
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both; a label marks that no cell carries both routes
Events that cannot both happen are mutually exclusive.
\[ P(I \mid F) = \frac{\textcolor{#b54708}{0}}{0.56} = \textcolor{#b54708}{0} \]
Check with the news test
Why: Fifth Street bars the Interstate
Trap
Figure (svg): A chance scale from 0 to 1 with a blue mark at the product 0.2464 and an orange mark at the counted overlap 0
\[ P(I \text{ AND } F) = \textcolor{#1f5fbf}{0.44} \times 0.56 \]
Apply the product rule
Why: 'unrelated' read as independence
\[ \textcolor{#1f5fbf}{0.44} \times 0.56 = \textcolor{#1f5fbf}{0.2464} \]
Multiply them
Why: fails: the overlap is 0
Figure (svg): The strip of 100 mornings, labelled I for Interstate and F for Fifth Street: 44 cells blue, 56 outlined, none in both
\[ P(I \mid F) = \textcolor{#b54708}{0} \ne \textcolor{#1f5fbf}{0.44} \]
Read the chance given F
Why: Fifth Street bars it
\[ \textcolor{#1f5fbf}{0.2464} \ne \textcolor{#b54708}{0} \]
Check product against count
Why: only zero could match
Worked example
Figure (svg): A chance scale from 0 to 1 with marks at P(I)P(F) and P(I AND F)
Suppose A and B are mutually exclusive, with P(A) > 0 and P(B) > 0.
\[ P(A \text{ AND } B) = \textcolor{#b54708}{0} \]
Write what exclusive means
Why: the hypothesis the claim starts from
\[ P(A)P(B) > \textcolor{#b54708}{0} \]
Multiply two positive chances
Why: nothing above zero can reach it
\[ 0.44 \times 0.56 = \textcolor{#1f5fbf}{0.2464} > \textcolor{#b54708}{0} \]
Check: set the two sides together
Why: no positive product can equal zero
Prediction
Figure (svg): A strip of 100 customers with 60 cells blue for card and 40 outlined for cash, none in both
Predict first
A shop's next customer pays by card (chance 0.6) or cash (chance 0.4), never both.
The clerk says the two are independent because one customer's payment cannot affect another's. Where is the flaw?
Correct: It is about these two events, not two customers
Why: The test asks what card news does to the cash chance for the SAME customer, and the answer is that it drops it to zero. The clerk answered a different question about two separate customers, which the shop's figures say nothing about.
Worked example
Figure (svg): A strip of 100 customers with 60 cells blue for card and 40 outlined for cash, none in both
K: pays by card · N: pays by cash
\[ P(N) = 0.4,\ \ P(K \text{ AND } N) = 0 \]
Write the two given chances
Why: one customer pays a single way
\[ P(N \mid K) = 0 \div \textcolor{#1f5fbf}{0.6} \]
Put the overlap over P(K)
Why: restricts to the card payers
\[ 0 \div \textcolor{#1f5fbf}{0.6} = 0 \]
Divide 0 by 0.6
Why: nothing like the 0.4 it started at
\[ P(K)P(N) = \textcolor{#1f5fbf}{0.6} \times 0.4 = 0.24 \]
Check with the product test
Why: 0.24 cannot match a zero overlap
Faded example
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both; a label marks that no cell carries both routes
Try It 3.12: P(I) = 0.44, P(F) = 0.56, P(I AND F) = 0.
Fill in the blanks
no cell is in both, so P(I OR F) = 0.44 + 0.56 = 1
Why: Because no morning lies in both blocks, adding the two counts never counts a cell twice: 44 + 56 = 100 of the 100 mornings. As chances that is 0.44 + 0.56 = 1, so Mark takes one of the two routes every morning.
Worked example
Figure (svg): A strip of 100 equally likely mornings: 44 cells filled blue for the Interstate and 56 outlined in heavy ink for Fifth Street, with no cell in both; a label marks that no cell carries both routes
\[ \textcolor{#1f5fbf}{44} + 56 = 100 \text{ cells} \]
Add the two blocks' cells
Why: no cell is counted twice here
\[ P(I \text{ OR } F) = \frac{100}{100} \]
Write that count as a chance
Why: either-route mornings over all
\[ \frac{100}{100} = 1 \]
Divide 100 by 100
Why: the whole strip is certain
\[ \textcolor{#1f5fbf}{0.44} + 0.56 = 1 \]
Check: add the two chances
Why: shares and counts agree
Section
Idea 4 of 4
Concept
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52
Example 3.4: a fair deck of 52 holds 13 spades.
Discussion prompt
Does what the first card was change the chance the second is a spade?
Answer:
It depends on one choice: whether the first card goes back into the deck.
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52
A: first card a spade · B: second card a spade
\[ P(A) = \textcolor{#1f5fbf}{13} \div 52 \]
Count spades in the deck
Why: all 52 cards equally likely
\[ \textcolor{#1f5fbf}{13} \div 52 = 0.25 \]
Divide 13 by 52
Why: a decimal to compare later
\[ P(B \mid A) = \textcolor{#1f5fbf}{13} \div 52 \]
Count spades in the replaced pool
Why: the shuffled-back card restored it
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52, B: second draw, replaced with 52
\[ P(B) = \textcolor{#1f5fbf}{13} \div 52 = 0.25 \]
Check B’s own chance
Why: the second draw alone gives the same
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52
\[ 52 - 1 = 51,\ \ \textcolor{#1f5fbf}{13} - 1 = \textcolor{#1f5fbf}{12} \]
Remove the drawn spade
Why: one card leaves both counts at once
\[ P(B \mid A) = \textcolor{#1f5fbf}{12} \div 51 \]
Count the spades left
Why: top and bottom both moved
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52, B: second draw, kept out with 51
\[ \textcolor{#1f5fbf}{12} \div 51 \approx 0.2353 \]
Divide 12 by 51
Why: a decimal to set against 0.25
\[ 0.25 - 0.2353 = 0.0147 \]
Subtract the two chances
Why: sizes the news's effect
\[ 0.2353 \times 51 \approx \textcolor{#1f5fbf}{12} \]
Check: rebuild the spades
Why: multiplying undoes dividing
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52
\[ P(B \mid A) = P(A \text{ AND } B) \div P(A) \]
Write the conditional rule
Why: the only link between both draws
\[ P(A) \cdot P(B \mid A) = P(A \text{ AND } B) \]
Multiply both sides by P(A)
Why: clears the fraction off the right
\[ P(A \text{ AND } B) = 0.25 \times \tfrac{\textcolor{#1f5fbf}{12}}{51} \]
Substitute the two draws' chances
Why: the second already carries the news
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52, B: second draw, kept out with 51
\[ P(A \text{ AND } B) \div 0.25 = \tfrac{\textcolor{#1f5fbf}{12}}{51} \]
Check: divide by the first draw
Why: returns the second draw's chance
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: A: first draw with 52, B: second draw, kept out with 51
\( {\tfrac{12}{51} \approx 0.2353}\;\;\Rightarrow\;\;\allowbreak {0.25 - 0.2353 = 0.0147} \)
\[ \textcolor{#1f5fbf}{1} \div 4 = 0.25 \]
Count one spade in four
Why: a made-up pool starting at 0.25
\[ 4 - 1 = 3,\ \ \textcolor{#1f5fbf}{1} - \textcolor{#1f5fbf}{1} = \textcolor{#1f5fbf}{0} \]
Draw the spade, recount both
Why: the only spade is gone for good
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: four cards, one spade with 4, spade drawn and kept out with 3
\[ \textcolor{#1f5fbf}{0} \div 3 = 0 \]
Count spades in the pool
Why: the share the second draw faces
\[ 0.25 - 0 = 0.25 > 0.0147 \]
Check: size the tiny pool's miss
Why: the same news moved it far more
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: first draw with 52, second draw, card replaced with 52
\[ P(\text{red}) = \textcolor{#1f5fbf}{26} \div 52 = 0.5 \]
Divide the red cards
Why: each card equally likely
\[ P(\text{no black}) = 0.5 \times 0.5 = 0.25 \]
Multiply the red draws
Why: replacement keeps them independent
\[ P(\text{some black}) = 1 - 0.25 = 0.75 \]
Take the complement
Why: some black is the opposite of none
\[ 0.75 + 0.25 = 1 \]
Check: add both outcomes
Why: some black or none, nothing else
Trap
Figure (svg): A bar of 52 cards with 13 spades blue, standing for a second draw from an unreduced pool
\[ P(\text{both spades}) = \tfrac{\textcolor{#1f5fbf}{13}}{52} \cdot \tfrac{\textcolor{#1f5fbf}{13}}{52} \]
Multiply unchanged draws
Why: the product rule, unchecked
\[ \tfrac{\textcolor{#1f5fbf}{13}}{52} \cdot \tfrac{\textcolor{#1f5fbf}{13}}{52} = \tfrac{1}{16} \]
Multiply the quarters
Why: fails: the card stayed out
Figure (svg): A shorter bar of 51 cards with 12 spades blue, the pool after a spade is kept out
\[ P(\text{both spades}) = \tfrac{\textcolor{#1f5fbf}{13}}{52} \cdot \tfrac{\textcolor{#1f5fbf}{12}}{51} \]
Chain the reduced pool
Why: draw two sees 51 cards
\[ \tfrac{\textcolor{#1f5fbf}{13}}{52} \cdot \tfrac{\textcolor{#1f5fbf}{12}}{51} = \tfrac{1}{17} < \tfrac{1}{16} \]
Check: multiply, compare
Why: losing a spade lowers it
Prediction
Figure (svg): Three lists of four card names, with the king of hearts appearing twice in the first list
Predict first
Try It 3.5: four cards are drawn. QS is the queen of spades, KS the king of spades, KH the king of hearts.
Which list could NOT have come from four draws without replacement?
Correct: KH, 7D, 6D, KH
Why: A deck holds one king of hearts. Once it is drawn and set aside it cannot appear again, so a list naming it twice needs the card put back. The other two lists name four different cards, which either method allows.
Worked example
Figure (svg): Three lists of four card names, with the king of hearts appearing twice in the first list
\[ \text{KH appears in places } 1 \text{ and } 4 \]
Find the list's repeat
Why: a deck holds one such king
\[ 52 - \textcolor{#1f5fbf}{1} = 51 \text{ cards left} \]
Set the first card aside
Why: without replacement it stays out
\[ P(\text{KH again}) = \textcolor{#1f5fbf}{0} \div 51 = \textcolor{#1f5fbf}{0} \]
Count the kings left
Why: chance zero: the list cannot happen
\[ P(\text{KH again, replaced}) = \textcolor{#1f5fbf}{1} \div 52 \]
Check the replaced method
Why: putting it back allows a repeat
Faded example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: first draw with 52
Twelve of the 52 cards are face cards. One is drawn and kept out.
Fill in the blanks
cards left = 51; face cards left = 11; P(face second | face first) = 0.216
Why: Setting one card aside leaves 52 − 1 = 51 cards, and since that card was a face card, 12 − 1 = 11 face cards remain. So the chance is 11 ÷ 51 ≈ 0.216, below the 12 ÷ 52 ≈ 0.231 the first draw had.
Worked example
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: first draw with 52
\[ 52 - 1 = 51,\ \ \textcolor{#1f5fbf}{12} - 1 = \textcolor{#1f5fbf}{11} \]
Take the drawn face card out
Why: it is set aside, not reshuffled
\[ P(\text{2nd face} \mid \text{1st face}) = \textcolor{#1f5fbf}{11} \div 51 \]
Count face cards left
Why: equally likely cards, reduced pool
Figure (svg): Bars drawn against one ruler, their lengths matching the pools: first draw with 52, second draw, card kept out with 51
\[ \textcolor{#1f5fbf}{11} \div 51 \approx 0.216,\ \ \textcolor{#1f5fbf}{12} \div 52 \approx 0.231 \]
Divide both
Why: decimals to set side by side
\[ 0.216 < 0.231 \]
Check the two chances' order
Why: losing a face card lowers it
Pattern
\[ \begin{aligned} P(A \mid B) &= P(A) \\ P(A \text{ AND } B) &= P(A)P(B) \\ P(A \text{ AND } B) &= P(A)P(B \mid A) \end{aligned} \]
Check
Figure (svg): The eight box cards with B4 and B5 in blue for G and B1 to B4 outlined in ink for H, B4 carrying both
Check your understanding
Example 3.11's box holds R1, R2, R3, B1 to B5. Let G = the number is above 3 and H = the card is a B numbered 1 to 4. Which verdict is right?
Answer: A
Why: G = {B4, B5} so P(G) = 2/8, H = {B1, B2, B3, B4} so P(H) = 4/8, and G AND H = {B4} so P(G AND H) = 1/8. The product (2/8)(4/8) = 1/8 matches, so the product test passes.
Worked example
Figure (svg): The eight box cards with B4 and B5 in blue for G and B1 to B4 outlined in ink for H, B4 carrying both
\[ P(G) = \textcolor{#1f5fbf}{2} \div 8,\ \ P(H) = 4 \div 8 \]
Count each event over the eight
Why: every card equally likely
\[ G \text{ AND } H = \{ \textcolor{#b54708}{B4} \} \]
Keep the cards in both
Why: only B4 is above 3 in H
Figure (svg): The eight cards with the R cards and B5 faded, B1 to B4 outlined and B4 filled orange
\[ (\textcolor{#1f5fbf}{2} \div 8) \times (4 \div 8) = \textcolor{#b54708}{1} \div 8 \]
Multiply the two chances
Why: the product test's prediction
\[ P(G \mid H) = 1 \div 4 = P(G) \]
Check with the news test
Why: one of H's four is above 3
Check
Figure (svg): Two bars of equal length, one for the 30 away-team fans with 20 filled blue for blue clothing and one for the 70 home fans with 5 filled blue
Check your understanding
Of 100 fans, 30 root for the away team, 25 wear blue, and 20 do both. Which pair of verdicts fits?
Answer: A
Why: The product test predicts 0.30 × 0.25 = 0.075, but the counted overlap is 0.20, so the events are dependent. An overlap of 0.20 is not zero, so they are not mutually exclusive.
Worked example
Figure (svg): Two bars of equal length, one for the 30 away-team fans with 20 filled blue for blue clothing and one for the 70 home fans with 5 filled blue
\[ \begin{aligned} 30 \div 100 &= 0.30 \\ \textcolor{#1f5fbf}{25} \div 100 &= \textcolor{#1f5fbf}{0.25} \\ \textcolor{#1f5fbf}{20} \div 100 &= 0.20 \end{aligned} \]
Divide each count by 100
Why: the three shares the tests need
\[ 0.30 \times \textcolor{#1f5fbf}{0.25} = 0.075 \]
Multiply the two chances
Why: the overlap independence would need
\[ 0.075 \ne 0.20,\ \ 0.20 \ne 0 \]
Compare the count with both
Why: dependent, and not exclusive
Figure (svg): Two bars of equal length, one for the 30 away-team fans with 20 filled blue for blue clothing and one for the 70 home fans with 5 filled blue, and a grey dashed line at the 0.25 share
\[ \frac{\textcolor{#1f5fbf}{20}}{30} \approx 0.67 \ne \textcolor{#1f5fbf}{0.25} \]
Check with the news test
Why: the book's 67% of away fans
Check
Figure (svg): A row of ten marbles labelled R1 to R6 and G1 to G4, with the four green marbles filled blue
Check your understanding
Try It 3.9: six red marbles numbered 1 to 6 and four green numbered 1 to 4. With G = green and O = odd-numbered, what is P(G AND O)?
Answer: A
Why: The marbles that are both green and odd-numbered are G1 and G3, so 2 of the 10 outcomes qualify and P(G AND O) = 0.2.
Worked example
Figure (svg): A row of ten marbles labelled R1 to R6 and G1 to G4, with the four green marbles filled blue
\[ \begin{aligned} G &= \{ \textcolor{#1f5fbf}{G1, G2, G3, G4} \} \\ O &= \{ R1, R3, R5, G1, G3 \} \end{aligned} \]
List both events
Why: ten outcomes, each equally likely
\[ P(G \text{ AND } O) = \textcolor{#b54708}{2} \div 10 = 0.2 \]
Count the marbles in both
Why: G1 and G3 are green and odd
Figure (svg): A row of ten marbles labelled R1 to R6 and G1 to G4, with the green marbles blue-filled, the odd-numbered outlined in ink, and G1 and G3 filled orange as both
\[ (\textcolor{#1f5fbf}{4} \div 10) \times (5 \div 10) = 0.2 \]
Check with the product test
Why: colour and parity are independent
Worked example
Figure (svg): A grid of 200 cells in 20 columns and 10 rows, one student each; the first 15 columns filled blue for the 150 who take English; the first 3 rows outlined in heavy ink for the 60 who take speech
\[ 150 \div 200 = 0.75,\ \ 60 \div 200 = 0.3 \]
Divide each count by 200
Why: the columns' share and the band's
\[ 0.75 \times 0.3 = \textcolor{#b54708}{0.225} \]
Multiply the two shares
Why: what independence would predict
Figure (svg): A grid of 200 cells in 20 columns and 10 rows, one student each; the first 15 columns filled blue for the 150 who take English; the first 3 rows outlined in heavy ink for the 60 who take speech; the 45 cells in both, 15 wide and 3 tall, filled orange
\[ \textcolor{#b54708}{0.225} = P(C \text{ AND } D) \]
Compare with the count
Why: the 45 cells counted earlier
\[ 15 \times 3 = 45 \text{ cells} \]
Check: the block is a rectangle
Why: a rectangle is what independence draws
Worked example
Figure (svg): A grid of 200 cells in 20 columns and 10 rows, one student each; the first 15 columns filled blue for the 150 who take English; the first 3 rows outlined in heavy ink for the 60 who take speech; the 45 cells in both, 15 wide and 3 tall, filled orange
\( {P(C \text{ AND } D) = 0.225} \)
\[ \textcolor{#b54708}{0.225} \ne 0 \]
Compare the overlap with zero
Why: the exclusivity test fails outright
\[ P(C \mid D) = \textcolor{#b54708}{0.225} \div 0.3 = \textcolor{#1f5fbf}{0.75} \]
Divide the overlap by P(D)
Why: English's chance among speech takers
\[ \textcolor{#1f5fbf}{0.75} = P(C) \]
Compare with P(C)
Why: independence wants these two equal
\[ P(D \mid C) = \textcolor{#b54708}{0.225} \div \textcolor{#1f5fbf}{0.75} = 0.3 \]
Check the reversed news
Why: speech keeps its own chance too
Recap
OpenStax Introductory Statistics 2e, §3.2 Independent and Mutually Exclusive Events §3.2, pp. 172-178 — Examples 3.4 to 3.12 and their Try Its trace back here
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