Count probabilities from pictures of equally likely outcomes, read probability as a long-run relative frequency, combine events with OR, AND and complements, and shrink the sample space to find conditional probabilities.
Subject: Statistics · 65 slides · applied lesson
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Title
Statistics · §3.1
Outcomes, events, OR, AND and news, each counted from a picture
Objectives
Concept
\[ \frac{u}{v} = \frac{u \div w}{v \div w} \]
Scale
Why: same value, cut into new parts
\[ \frac{u}{w} + \frac{v}{w} = \frac{u + v}{w} \]
Like fractions
Why: parts of one size add up
\[ \frac{w}{w} = 1 \]
Whole
Why: w equal parts make one
\[ \frac{u / w}{v / w} = \frac{u}{v} \]
Shared bottoms
Why: the common w cancels
\[ u \times v \text{ cells} \]
Grid count
Why: rows times columns, each pair once
\[ \text{count} \div \text{total} \]
Relative frequency
Why: a value's data share
\[ u\% = u \div 100 \]
Percent
Why: parts per hundred
\[ 2 \div 3 \approx 0.67 \]
Round
Why: ≈ marks a cut-off decimal
Prediction
Predict first
A friend peeks at a tossed dime and nickel: 'at least one is heads.'
Is both heads now a 50-50?
Correct: No: less than one half
Why: Calling 'the other coin' a 50-50 pretends we know which coin showed heads, and the friend never said. The deck builds a way to count chances with news, then comes back for the exact answer.
Worked example
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells still empty
H: heads, T: tails. HT: dime H, nickel T.
\[ \text{dime: } H,\ T \]
List the dime's faces
Why: they label the grid's rows
\[ \text{nickel: } H,\ T \]
List the nickel's faces
Why: columns, each pairing with both
\[ 2 \times 2 = 4 \]
Count rows times columns
Why: each dime face meets each nickel
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells HH, HT, TH, TT
\[ HH,\ HT,\ TH,\ TT \]
Check: read every cell aloud
Why: four pairings, none repeated or missing
Section
Idea 1 of 4
Concept
Figure (svg): The six faces of one fair die as plain cells 1 to 6
Roll one fair die: a 5 or a 6 moves you ahead.
Discussion prompt
What single number could say how likely a move ahead is, before you roll?
Answer:
List every face the die can show, then weigh the winning faces against all of them.
Worked example
Figure (svg): Two dice as cell strips: a six-sided die 1 to 6 with 5 and 6 shaded blue; the ten-sided die's strip not drawn yet
\[ \text{six-sided: } \textcolor{#1f5fbf}{5, 6} \to \textcolor{#1f5fbf}{2} \text{ wins} \]
Count the six-sided die's wins
Why: the obvious measure of chance
\[ \text{ten-sided: } \textcolor{#1f5fbf}{9, 10} \to \textcolor{#1f5fbf}{2} \text{ wins} \]
Count a ten-sided die's top two
Why: can counts compare two dice?
Figure (svg): Two dice as cell strips: a six-sided die 1 to 6 with 5 and 6 shaded blue; a ten-sided die 1 to 10 with 9 and 10 shaded blue
\[ \textcolor{#1f5fbf}{2} = \textcolor{#1f5fbf}{2} \]
Compare the two win counts
Why: the test counts must pass
\[ 6 - \textcolor{#1f5fbf}{2} = 4,\ \ 10 - \textcolor{#1f5fbf}{2} = 8 \]
Check: count each die's losers
Why: same wins, spread over more faces
Worked example
Figure (svg): The six faces of one fair die as plain cells 1 to 6
Experiment: one roll. Outcome: the face that shows.
\[ S = \{1, 2, 3, 4, 5, 6\} \]
Collect every outcome as S
Why: a sample space misses no result
\[ E = \{\textcolor{#1f5fbf}{5}, \textcolor{#1f5fbf}{6}\} \]
Collect the winning outcomes as E
Why: an event: any set of outcomes
Figure (svg): The six faces of one fair die as cells 1 to 6, with 5 and 6 shaded blue as the event E
\[ 6 \text{ in } S,\ \ \textcolor{#1f5fbf}{2} \text{ in } E \]
Count both sets
Why: weighing needs whole and part
\[ \{1, 2, 3, 4\}\colon\ 4 + \textcolor{#1f5fbf}{2} = 6 \]
Check: add the losers to E
Why: no face is missed or doubled
Worked example
Figure (svg): The six faces of one fair die as cells 1 to 6, with 5 and 6 shaded blue as the event E
\[ P(S) = 1 \]
Score S's chance, P(S), as 1
Why: some face must show
\[ 1 \div 6 = \frac{1}{6} \]
Share 1 among the 6 faces
Why: fair: equally likely faces, equal shares
Figure (svg): The six faces of one fair die, each labelled with its share 1/6
\[ P(E) = \textcolor{#1f5fbf}{\frac{1}{6} + \frac{1}{6}} \]
Collect face 5's and 6's shares
Why: an event gathers its outcomes' chances
Figure (svg): The six faces of one fair die, each labelled with its share 1/6, with 5 and 6 shaded blue
\[ \textcolor{#1f5fbf}{\frac{1}{6} + \frac{1}{6}} = \frac{\textcolor{#1f5fbf}{2}}{6} \]
Add like fractions
Why: the win needs one single number
\[ \frac{\textcolor{#1f5fbf}{2}}{6} + \frac{4}{6} = 1 \]
Check: add the four losing shares
Why: E and the losers rebuild certainty
Worked example
Figure (svg): The six faces of one fair die as cells 1 to 6, with 5 and 6 shaded blue as the event E
\( {P(S) = 1}\;\;\Rightarrow\;\;\allowbreak {\dfrac{1}{6} \text{ per face}}\;\;\Rightarrow\;\;\allowbreak {P(E) = \dfrac{\textcolor{#1f5fbf}{2}}{6}} \)
\[ P(E) = \frac{\textcolor{#1f5fbf}{2 \text{ in } E}}{6 \text{ in } S} \]
Read 2 and 6 as counts
Why: each E face brings one sixth
Figure (svg): The six faces of one fair die with 5 and 6 shaded blue; below: 2 blue cells in E, 6 cells in S
\[ P(A) = \frac{\textcolor{#1f5fbf}{\text{outcomes in } A}}{\text{outcomes in } S} \]
Replace E by any event A
Why: any S whose outcomes share equally
Equally likely: every outcome has the same chance.
\[ A = S\colon\ \frac{6}{6} = 1 \]
Check: let A be all of S
Why: a certain event must score 1
Worked example
Figure (svg): The die strip with 5 and 6 shaded blue above an empty chance scale marked 0 and 1
\[ \text{face } 7\colon\ 0 \text{ outcomes} \]
Count the faces showing 7
Why: an event no roll can give
\[ \frac{0}{6} = \textcolor{#6b7280}{0} \]
Divide 0 by 6
Why: an impossible event scores 0
Figure (svg): The die strip above a chance scale from 0 to 1, with a grey dot at 0 labelled never
\[ \text{faces 1 to 6: } \frac{6}{6} = \textcolor{#6b7280}{1} \]
Divide all 6 faces by 6
Why: a certain event scores 1
Figure (svg): The chance scale with grey dots at 0, never, and 1, always
\[ \textcolor{#6b7280}{0} \le P(A) \le \textcolor{#6b7280}{1} \]
Place every event between
Why: counts run from none to all
Figure (svg): The chance scale with the whole stretch from 0 to 1 shaded grey between never and always
\[ \textcolor{#6b7280}{0} \le \frac{\textcolor{#1f5fbf}{2}}{6} \le \textcolor{#6b7280}{1} \]
Check the win chance sits inside
Why: a real event must land inside
Figure (svg): The shaded chance scale with a blue dot at 2/6, a third of the way from 0 to 1
Worked example
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells HH, HT, TH, TT
\[ S = \{HH, HT, TH, TT\} \]
Read S off the grid
Why: fair coins, four equal cells
\[ A = \{\textcolor{#1f5fbf}{HT}, \textcolor{#1f5fbf}{TH}\} \]
Shade the one-head cells
Why: marks the cells to count
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells HH, HT, TH, TT
\[ P(A) = \frac{\textcolor{#1f5fbf}{2}}{4} \]
Divide A's count by 4
Why: fair coins: equal cells
\[ \frac{\textcolor{#1f5fbf}{2}}{4} = 0.5 \]
Divide 2 by 4
Why: 0.5 reads as even odds
\[ HH,\ TT\colon\ 2 \text{ unshaded of } 4 \]
Check: count cells outside A
Why: 0.5 must mean half the cells
Trap
\[ \{0 \text{ H},\ 1 \text{ H},\ 2 \text{ H}\} \]
List head counts as outcomes
Why: a tempting shorter list
\[ P(1 \text{ H}) = \frac{1}{3} \]
Give each head count 1/3
Why: treats them as equally likely
\[ \frac{1}{3} \ne \frac{2}{4} \]
Compare with the grid
Why: fails: one head fills two cells
Figure (svg): The four coin cells HH, HT, TH, TT in a row, with HT and TH shaded blue
\[ P(1 \text{ H}) = \frac{\textcolor{#1f5fbf}{2}}{4} \]
Count one-head cells over 4
Why: only cells are equally likely
\[ \frac{1}{4} + \frac{\textcolor{#1f5fbf}{2}}{4} + \frac{1}{4} = 1 \]
Check: shares of 2, 1, 0 heads
Why: the grid's cells, each counted once
Prediction
Predict first
Tomorrow it either rains or stays dry: two outcomes.
Does that make the chance of rain one half?
Correct: No: the two outcomes need not be equally likely
Why: The counting rule shares 1 equally only among equally likely outcomes. Rain and dry are two outcomes, but nothing makes them equal: a spinner with one rain slice and three dry slices also has just two outcomes.
Worked example
Figure (svg): A spinner cut into two halves labelled rain and dry: the naive two-outcome view
Spinner model: 4 equal slices, 1 rain.
\[ S = \{\text{rain},\ \text{dry}\} \]
List the two outcomes
Why: lists need no equal chances
\[ \text{rain: } \textcolor{#1f5fbf}{1} \text{ slice},\ \text{dry: } 3 \text{ slices} \]
Sort the 4 slices
Why: slices are the equal units
Figure (svg): A spinner cut into four equal slices: one rain slice shaded blue and three dry slices
\[ P(\text{rain}) = \frac{\textcolor{#1f5fbf}{1}}{4} \]
Divide rain slices by 4
Why: equal units license counting
\[ \frac{\textcolor{#1f5fbf}{1}}{4} + \frac{3}{4} = 1 \]
Check: add the dry share
Why: every slice is rain or dry
Faded example
Figure (svg): The numbers 1 to 10 as plain cells
Try It 3.2: pick a number from 1 to 10 at random.
Fill in the blanks
A = more than 6 holds 7, 8, 9, 10; P(A) = 4/10
Why: S has the 10 numbers. More than 6 means 7, 8, 9 and 10, four outcomes, so P(A) = 4/10 = 0.4.
Worked example
Figure (svg): The numbers 1 to 10 as plain cells
\[ S = \{1, 2, \ldots, 10\} \]
Take the ten numbers as S
Why: each pick equally likely
\[ A = \{\textcolor{#1f5fbf}{7, 8, 9, 10}\} \]
Shade the numbers above 6
Why: 'more than' excludes 6
Figure (svg): The numbers 1 to 10 as cells with 7, 8, 9 and 10 shaded blue
\[ P(A) = \frac{\textcolor{#1f5fbf}{4}}{10} \]
Divide A's count by 10
Why: fair picks license counting
\[ 6 + \textcolor{#1f5fbf}{4} = 10 \]
Check: add the 6 unshaded cells
Why: A and the rest fill S
Section
Idea 2 of 4
Concept
Figure (svg): A bar of the Belgian euro's 250 trials split into 56% heads (blue) and 44% tails
Book: statistics students tested the coin 250 times.
Discussion prompt
Does 56% heads in 250 trials prove the coin is not fair?
Answer:
First see how far a coin we know is fair strays in runs of different lengths.
Worked example
Figure (svg): The first 10 simulated flips of a fair coin as cells: T H H H H H H H T T, with the 7 heads shaded blue
A computer flipped a fair coin: P(H) = 0.5.
\[ \textcolor{#1f5fbf}{7} \text{ H in } 10 \]
Count heads in 10 flips
Why: a quick fairness test
\[ \frac{\textcolor{#1f5fbf}{7}}{10} = \textcolor{#1f5fbf}{0.7} \]
Divide heads by 10 flips
Why: shares face P(H) directly
\[ \textcolor{#1f5fbf}{0.7} - \textcolor{#6b7280}{0.5} = \textcolor{#1f5fbf}{0.2} \]
Subtract the true chance
Why: fair coins' misses set the bar
\[ \frac{\textcolor{#1f5fbf}{7}}{10} - \frac{5}{10} = \frac{\textcolor{#1f5fbf}{2}}{10} \]
Check: redo it in tenths
Why: 2 tenths is the same 0.2
Worked example
Figure (svg): Running share of heads against flips on a ten-fold axis with the first point 0.7 at 10 flips and a grey line at P(H) 0.5
Simulated heads: 43 in 100, 509 in 1000, 5023 in 10000.
\[ \frac{\textcolor{#1f5fbf}{43}}{100} = \textcolor{#1f5fbf}{0.43} \]
Divide 43 by 100 flips
Why: shares compare any run lengths
Figure (svg): Running share of heads against flips on a ten-fold axis, drawn to 100 flips, with 0.7 at 10 and 0.43 at 100, and a grey line at P(H) 0.5
\[ \frac{\textcolor{#1f5fbf}{509}}{1000} = \textcolor{#1f5fbf}{0.509} \]
Divide 509 by 1000
Why: was the dip a trend?
Figure (svg): The running share drawn to 1000 flips, with points 0.7, 0.43 and 0.509
\[ \frac{\textcolor{#1f5fbf}{5023}}{10000} = \textcolor{#1f5fbf}{0.5023} \]
Divide 5023 by 10000
Why: does the settling continue?
Figure (svg): The running share drawn to 10000 flips, with points 0.7, 0.43, 0.509 and 0.5023 hugging the grey line at 0.5
\[ \textcolor{#1f5fbf}{0.509} \times 1000 = \textcolor{#1f5fbf}{509},\ \ \textcolor{#1f5fbf}{0.5023} \times 10000 = \textcolor{#1f5fbf}{5023} \]
Check: undo the two longest divisions
Why: recovers the simulated head counts
Worked example
Figure (svg): The running share of heads drawn to 10000 flips, with points 0.7, 0.43, 0.509 and 0.5023, a grey line at P(H) 0.5, and a blue miss bar from 0.7 down to 0.5
\( {\textcolor{#1f5fbf}{0.2} \text{ miss}},\ \allowbreak \allowbreak {\textcolor{#1f5fbf}{0.43}},\ \allowbreak \allowbreak {\textcolor{#1f5fbf}{0.509}},\ \allowbreak \allowbreak {\textcolor{#1f5fbf}{0.5023}} \)
\[ \textcolor{#6b7280}{0.5} - \textcolor{#1f5fbf}{0.43} = \textcolor{#1f5fbf}{0.07} \]
Find the 100-flip miss
Why: larger minus smaller: no negatives
Figure (svg): The running share with blue miss bars from 0.7 and from 0.43 to the grey line at 0.5
\[ \textcolor{#1f5fbf}{0.509} - \textcolor{#6b7280}{0.5} = \textcolor{#1f5fbf}{0.009} \]
Find the 1000-flip miss
Why: did 10× the flips help?
Figure (svg): The running share with miss bars at 10, 100 and a short bar at 1000 flips
\[ \textcolor{#1f5fbf}{0.5023} - \textcolor{#6b7280}{0.5} = \textcolor{#1f5fbf}{0.0023} \]
Find the 10000-flip miss
Why: did another 10× help?
Figure (svg): The running share with miss bars at all four runs, the 10000-flip bar barely visible
\[ \textcolor{#1f5fbf}{0.2} > \textcolor{#1f5fbf}{0.07} > \textcolor{#1f5fbf}{0.009} > \textcolor{#1f5fbf}{0.0023} \]
Check: order the misses
Why: shrinking misses are the law's claim
Worked example
Figure (svg): A bar of the Belgian euro's 250 trials split into 56% heads (blue) and 44% tails
\[ \textcolor{#1f5fbf}{56\%} = \textcolor{#1f5fbf}{0.56} \]
Write 56% as a decimal
Why: a decimal can scale 250
\[ \textcolor{#1f5fbf}{0.56} \times 250 = \textcolor{#1f5fbf}{140} \]
Take 0.56 of 250 trials
Why: counts show the lopsidedness
Figure (svg): A bar of the euro's 250 trials split into 140 heads (blue) and 110 tails
\[ \textcolor{#1f5fbf}{0.56} - \textcolor{#6b7280}{0.5} = \textcolor{#1f5fbf}{0.06} \]
Subtract the fair chance
Why: a miss compares with fair coins
\[ \textcolor{#1f5fbf}{0.06} \text{ vs } \textcolor{#1f5fbf}{0.07} \text{ at } 100 \]
Set it beside our fair coin's miss
Why: but 250 trials should miss less
\[ \textcolor{#1f5fbf}{140} \div 250 = \textcolor{#1f5fbf}{0.56} \]
Check: divide heads by 250
Why: undoes the count
Trap
\[ P(\text{T next}) > 0.5 \]
Expect tails after 7 H, 3 T
Why: assumes the coin remembers
\[ P(\text{T next}) = 0.5 \]
Compare with a fair coin
Why: fails: no memory, every flip 50-50
\[ \text{expect } 990 \div 2 = 495 \]
Expect about half of 990 new flips
Why: each new flip is still fair
\[ \text{expect } \textcolor{#1f5fbf}{7} + 495 = \textcolor{#1f5fbf}{502} \]
Add the early 7 heads
Why: kept, just outnumbered
\[ 10 + 990 = 1000 \]
Total the flips
Why: early ones count too
\[ \textcolor{#1f5fbf}{502} \div 1000 = \textcolor{#1f5fbf}{0.502} \]
Check: expected share after 1000
Why: near 0.5, with no catch-up
Prediction
Predict first
Hospitals with 15 and 45 births a day; about half are boys.
Which records more days with over 60% boys?
Correct: The small hospital
Why: By the law of large numbers, a day's share of boys stays nearer one half when there are more births. The small hospital's days stray further from half, so they cross 60% more often.
Worked example
Figure (svg): Two rows of birth cells: 15 for the small hospital and 45, in three rows of 15, for the large one
\[ 0.6 \times 15 = \textcolor{#6b7280}{9} \]
Find 60% of 15 births
Why: more boys than this pass 60%
Figure (svg): Two rows of birth cells, 15 for the small hospital and 45 for the large one, with a grey line after the 9th cell of the small row
\[ 0.6 \times 45 = \textcolor{#6b7280}{27} \]
Find 60% of 45 births
Why: large days need the same bar
Figure (svg): The birth rows with grey cut-off lines after cell 9 of 15 and cell 27 of 45, labelled 60%: 9 and 60%: 27
\[ \text{one birth: } \frac{\textcolor{#1f5fbf}{1}}{15},\ \ \frac{\textcolor{#1f5fbf}{1}}{45} \]
Share one birth in each day
Why: step size sets the straying
Figure (svg): The birth rows with both cut-off lines, and the first cell of each hospital's row shaded blue as one birth
\[ \frac{\textcolor{#1f5fbf}{1}}{45} + \frac{\textcolor{#1f5fbf}{1}}{45} + \frac{\textcolor{#1f5fbf}{1}}{45} = \frac{\textcolor{#1f5fbf}{3}}{45} \]
Add three large-day steps
Why: tests the 3× claim against 1/15
Figure (svg): The birth rows with cut-off lines, one small-day cell and three large-day cells shaded blue
\[ \frac{\textcolor{#1f5fbf}{3} \div 3}{45 \div 3} = \frac{\textcolor{#1f5fbf}{1}}{15} \]
Check: reduce 3/45 by 3
Why: three large steps make one small
Faded example
Figure (svg): The six faces of one fair die as cells 1 to 6, with 5 and 6 shaded blue as the event E
E: rolling 5 or 6, with P(E) = 2/6.
Fill in the blanks
Rolls in E to expect ≈ 200; a run with 230 such rolls has relative frequency ≈ 0.383
Why: 600 ÷ 6 = 100 rolls per face, and E has 2 faces: about 200. A run with 230 gives 230 ÷ 600 ≈ 0.383, a little above the long-run share.
Worked example
Figure (svg): The six faces of one fair die as cells 1 to 6, with 5 and 6 shaded blue as the event E
\[ 600 \div 6 = 100 \]
Split 600 rolls among 6 faces
Why: equal faces, equal rolls
Figure (svg): A bar of 600 rolls cut into six equal parts, each labelled 100
\[ 2 \times 100 = \textcolor{#1f5fbf}{200} \]
Take the two faces in E
Why: 5 and 6 expect 100 each
Figure (svg): The 600-roll bar with its last two parts merged and shaded blue, labelled E: 200
\[ \textcolor{#1f5fbf}{230} \div 600 \approx \textcolor{#1f5fbf}{0.383} \]
Divide the run's count by 600
Why: puts the run on P's scale
Figure (svg): Two bars over 600 rolls: expected, with E: 200 shaded blue at the right end; one run, with E: 230 shaded blue
\[ \textcolor{#6b7280}{2} \div \textcolor{#6b7280}{6} \approx \textcolor{#6b7280}{0.333} \]
Write P(E) as a decimal
Why: a decimal to set beside 0.383
Figure (svg): The two 600-roll bars, with a grey dashed line on the run's bar where the expected E: 200 begins, labelled expected E starts, so the run's E: 230 reaches past it
\[ \textcolor{#1f5fbf}{0.383} \times 600 \approx \textcolor{#1f5fbf}{230},\ \ \textcolor{#6b7280}{0.333} \times 600 \approx \textcolor{#1f5fbf}{200} \]
Check: scale both shares by 600
Why: undoes both divisions
Section
Idea 3 of 4
Concept
Figure (svg): The raffle's tickets 1 to 19 as plain cells in two rows
Example 3.1, as a raffle: one ticket from 1 to 19.
Discussion prompt
A prize goes to every ticket that is even or above 13. How many of the 19 tickets win?
Answer:
Mark the even tickets and the tickets above 13, then count each winning ticket once.
Worked example
Figure (svg): Tickets 1 to 19 in two blocks of numbered columns, no rows marked yet
A: even tickets. B: tickets above 13.
\[ A = \{2, 4, \ldots, 18\}\colon\ 9 \]
Count the even tickets
Why: plan: tally each group
Figure (svg): Tickets 1 to 19 in two blocks with row A marking the even tickets 2 to 18 by dots
\[ B = \{14, 15, \ldots, 19\}\colon\ 6 \]
Count the tickets above 13
Why: adding needs both tallies
Figure (svg): The ticket blocks with row A marking even tickets and row B marking tickets 14 to 19
\[ 9 + 6 = 15 \]
Add the two counts
Why: seems to count winners
\[ 14, 16, 18\colon\ \text{marked twice} \]
Check: look for double marks
Why: 15 counted these three twice
Figure (svg): The ticket blocks with rows A and B and the columns 14, 16 and 18, marked in both rows, boxed
Worked example
Figure (svg): Tickets 1 to 19 in two blocks with row A marking even tickets and row B marking tickets 14 to 19
\[ A \text{ AND } B = \{\textcolor{#1f5fbf}{14, 16, 18}\} \]
Keep columns marked twice
Why: AND needs both conditions
Figure (svg): The ticket blocks with rows A and B and a blue AND row shading tickets 14, 16 and 18
\[ A \text{ OR } B\colon\ \textcolor{#1f5fbf}{12} \text{ tickets} \]
Keep columns with any mark
Why: OR accepts either, or both
Figure (svg): The ticket blocks with rows A and B and a blue OR row shading tickets 2, 4, 6, 8, 10, 12 and 14 to 19
Book symbols: A ∪ B is A OR B; A ∩ B is A AND B.
\[ 15 - \textcolor{#1f5fbf}{3} = \textcolor{#1f5fbf}{12} \]
Check: remove the double count
Why: the attempt counted 3 twice
Worked example
Figure (svg): Tickets 1 to 19 in two blocks with rows A and B marked
\[ P(A) = \frac{9}{19},\ \ P(B) = \frac{6}{19} \]
Divide both row counts by 19
Why: fair draws make tickets equally likely
\[ P(A \text{ AND } B) = \frac{\textcolor{#1f5fbf}{3}}{19} \]
Divide the AND count by 19
Why: each overlap ticket carries one nineteenth
Figure (svg): The ticket blocks with a blue AND row shading 14, 16 and 18
\[ P(A \text{ OR } B) = \frac{\textcolor{#1f5fbf}{12}}{19} \]
Divide the OR count by 19
Why: counting covers combined events too
Figure (svg): The ticket blocks with a blue OR row shading the 12 winning tickets
\[ \frac{\textcolor{#1f5fbf}{3}}{19} \le \frac{9}{19} \le \frac{\textcolor{#1f5fbf}{12}}{19} \]
Check: order AND, A, OR
Why: AND inside A, A inside OR
Concept
Figure (svg): The raffle's tickets 1 to 19 with the 9 even tickets shaded blue
A: the 9 even tickets. A′ (A prime): every ticket not in A.
Discussion prompt
How could A's count of 9 give P(A′) without listing A′?
Answer:
Every ticket is even or not, never both: the rest of the 19 are exactly A′.
Worked example
Figure (svg): Tickets 1 to 19 in blocks of seven with row A marking the 9 even tickets by dots; the A′ row not drawn yet
\[ 19 - 9 = \textcolor{#1f5fbf}{10} \text{ in } A' \]
Take A's 9 from 19
Why: each ticket is even or not
Figure (svg): Tickets 1 to 19 in blocks of seven: row A marks the 9 even tickets with dots, and a blue row A′ shades the 10 odd tickets
\[ \textcolor{#1f5fbf}{P(A')} = \frac{\textcolor{#1f5fbf}{10}}{19} \]
Divide A′'s count by 19
Why: equally likely tickets license counting
Figure (svg): The A and A′ rows with the note A prime: 10 of 19 tickets
\[ A' = \{\textcolor{#1f5fbf}{1, 3, 5, \ldots, 19}\}\colon\ \textcolor{#1f5fbf}{10} \]
Check: list the odd tickets
Why: a count made without subtracting
Worked example
Figure (svg): Tickets 1 to 19 in blocks of seven: row A marks the 9 even tickets with dots, and a blue row A′ shades the 10 odd tickets
\( {P(A) = \dfrac{9}{19}},\quad\allowbreak \allowbreak {\textcolor{#1f5fbf}{P(A')} = \dfrac{\textcolor{#1f5fbf}{10}}{19}} \)
\[ \frac{9}{19} + \frac{\textcolor{#1f5fbf}{10}}{19} = \frac{19}{19} \]
Add the two chances
Why: like fractions add their tops
Figure (svg): The A and A′ rows with the note 9 dots plus 10 blue make 19
\[ \frac{19}{19} = 1 \]
Use the whole rule
Why: all 19 tickets are certain
\[ P(A) + \textcolor{#1f5fbf}{P(A')} = 1 \]
Name each chance
Why: each outcome sits in exactly one
\[ \text{each of 19 columns: one mark} \]
Check: count the marks per column
Why: no ticket twice, none missing
Figure (svg): The A and A′ rows with every ticket's column boxed, each holding exactly one mark
Worked example
Figure (svg): The raffle's tickets 1 to 19 with the 10 odd tickets, the event A prime, shaded blue
\( {P(A) + \textcolor{#1f5fbf}{P(A')} = 1} \)
\[ \textcolor{#1f5fbf}{P(A')} = 1 - P(A) \]
Subtract P(A) from both sides
Why: gives A′ from A alone
\[ C = S,\ \ \textcolor{#1f5fbf}{P(C)} = 1 \]
Let C be every ticket
Why: an edge: nothing left outside
Figure (svg): All 19 raffle tickets shaded blue as event C, with the note outside C: none
\[ P(C') = 1 - \textcolor{#1f5fbf}{1} \]
Put in P(C) = 1
Why: any event must obey it
\[ 1 - \textcolor{#1f5fbf}{1} = 0 \]
Subtract
Why: gives a value counting can test
\[ C' = \{\,\}\colon\ \frac{0}{19} = 0 \]
Check: count tickets in C′
Why: counting agrees: none of 19
Trap
\[ A = \{2, 4, 6\},\ B = \{1, 2, 3\} \]
List the even and under-4 faces
Why: their overlap tests OR
\[ \text{just one: } \{1, 3, 4, 6\} \]
Keep faces in one event only
Why: reads OR as either, not both
\[ P(\text{just one}) = \frac{4}{6} \]
Divide 4 faces by 6
Why: fails: OR keeps shared face 2
\[ A \text{ OR } B = \{\textcolor{#1f5fbf}{1, 2, 3, 4, 6}\} \]
Keep faces in A, B or both
Why: overlap included, counted once
\[ P(A \text{ OR } B) = \frac{\textcolor{#1f5fbf}{5}}{6} \]
Divide the 5 faces by 6
Why: fair faces make counting valid
\[ \frac{\textcolor{#1f5fbf}{5}}{6} + \frac{1}{6} = 1 \]
Check: add face 5, in neither
Why: OR and its complement fill S
Prediction
Predict first
A and B are two events of the same experiment.
Can P(A OR B) ever be smaller than P(A)?
Correct: No: A OR B contains every outcome of A
Why: Every outcome in A is also in A OR B, so A OR B holds at least A's outcomes. An unlikely B or an overlap can leave the chance unchanged, never cut it.
Worked example
Figure (svg): Faces 1 to 6 with row A marking 2, 4, 6 and row B marking only 6
\[ A = \{2, 4, 6\},\ B = \{6\} \]
Pick B inside A
Why: B then adds the least possible
\[ A \text{ OR } B = \{\textcolor{#1f5fbf}{2, 4, 6}\} \]
Keep faces in either event
Why: nothing in A can drop out
Figure (svg): Faces 1 to 6 with row A marking 2, 4, 6 and row B marking only 6, and a blue OR row shading 2, 4 and 6
\[ P(A \text{ OR } B) = \frac{\textcolor{#1f5fbf}{3}}{6} \]
Divide the 3 faces by 6
Why: can OR fall below P(A)?
\[ \textcolor{#1f5fbf}{2, 4, 6}\colon\ \text{each in the OR row} \]
Check: find A's faces in the OR row
Why: none of A's outcomes was lost
Faded example
Figure (svg): The six faces of one fair die as plain cells 1 to 6
Example 3.2: A = even face, B = face under 4.
Fill in the blanks
B′ holds faces 4, 5, 6; A OR B′ holds faces 2, 4, 5, 6; P(A OR B′) = 4/6
Why: B′ is the faces not under 4: 4, 5, 6. A OR B′ keeps faces in A (2, 4, 6) or B′ (4, 5, 6) or both: 2, 4, 5, 6. That is 4 of the 6 faces.
Worked example
Figure (svg): Faces 1 to 6 with row A marking 2, 4, 6; the B′ row not drawn yet
\[ B' = \{4, 5, 6\} \]
Take the faces not under 4
Why: B′ is everything B leaves out
Figure (svg): Faces 1 to 6 with row A marking 2, 4, 6 and row B′ marking 4, 5, 6
\[ A \text{ OR } B' = \{\textcolor{#1f5fbf}{2, 4, 5, 6}\} \]
Keep faces in either row
Why: shared faces 4, 6 count once
Figure (svg): Faces 1 to 6 with row A marking 2, 4, 6 and row B′ marking 4, 5, 6, and a blue OR row shading 2, 4, 5 and 6
\[ P(A \text{ OR } B') = \frac{\textcolor{#1f5fbf}{4}}{6} \]
Divide the 4 faces by 6
Why: fair faces share 1 equally
\[ \{1, 3\}\colon\ 2 \text{ faces in neither} \]
Count the unmarked faces
Why: the leftovers let us check
\[ \frac{\textcolor{#1f5fbf}{4}}{6} + \frac{2}{6} = 1 \]
Check: add the neither share
Why: event and rest must fill S
Section
Idea 4 of 4
Concept
Figure (svg): The six faces of one fair die with 2 and 3 shaded blue as event A
Book example: A = the face is 2 or 3.
Discussion prompt
A had 2 faces of 6. Once you hear 'even', how likely is A now?
Answer:
The news rules out the odd faces; count A's faces among those that remain.
Worked example
Figure (svg): The six faces of one fair die with 2 and 3 shaded blue as event A
\[ P(A) = \frac{\textcolor{#1f5fbf}{2}}{6} \]
Count A's faces over all 6
Why: a baseline to judge the news
\[ \textcolor{#6b7280}{1, 3, 5} \text{ cannot show} \]
Apply the news: the roll is even
Why: odd faces are now impossible
Figure (svg): The six faces with odd faces 1, 3 and 5 greyed and dashed as ruled out; face 2 shaded blue
\[ \textcolor{#6b7280}{3} \text{ is odd} \]
Recheck A's two faces
Why: 2/6 still counts a ruled-out face
\[ \{\textcolor{#1f5fbf}{2}, \textcolor{#6b7280}{3}\} \text{ vs } \{2, 4, 6\}\colon\ \text{only } \textcolor{#1f5fbf}{2} \]
Check: which of A's faces survive
Why: counting 3 overstates A's chance
Worked example
Figure (svg): The six faces with odd faces greyed out as ruled out and face 2 shaded blue
B: an even roll. P(A | B): A's chance given B.
\[ B = \{2, 4, 6\} \]
Make B the new sample space
Why: only these faces can have shown
Figure (svg): The six faces with 2, 4 and 6 heavily outlined as B and odd faces greyed out
\[ A \text{ AND } B = \{\textcolor{#1f5fbf}{2}\} \]
Keep A's faces inside B
Why: A's outcomes still in play
Figure (svg): The six faces with 2, 4 and 6 heavily outlined as the new sample space B, face 2 also shaded blue, and odd faces greyed out
\[ P(A \mid B) = \frac{\textcolor{#1f5fbf}{1}}{3} \]
Divide by B's 3 faces, not 6
Why: even faces stay equally likely
\[ \frac{\textcolor{#1f5fbf}{1}}{3} + \frac{2}{3} = 1 \]
Check: add faces 4 and 6
Why: B's faces make the new whole
Worked example
Figure (svg): The six faces with 2, 4 and 6 heavily outlined as the new sample space B, face 2 also shaded blue, and odd faces greyed out
\( {P(A \mid B) = \dfrac{\textcolor{#1f5fbf}{1}}{3}} \)
\[ \frac{\textcolor{#1f5fbf}{1}}{3} = \frac{\textcolor{#1f5fbf}{1} \div 6}{3 \div 6} \]
Divide top and bottom by 6
Why: brings back S's six faces
Figure (svg): All six faces back in plain ink as S, with 2, 4 and 6 still heavily outlined as B and face 2 shaded blue
\[ \textcolor{#1f5fbf}{1} \div 6 = \textcolor{#1f5fbf}{P(A \text{ AND } B)} \]
Read the top as a chance
Why: count over 6 is a chance
Figure (svg): All six faces back in plain ink as S, with 2, 4 and 6 still heavily outlined as B and face 2 shaded blue
\[ 3 \div 6 = P(B) \]
Read the bottom as a chance
Why: the bottom needs a chance too
Figure (svg): All six faces back in plain ink as S, with 2, 4 and 6 still heavily outlined as B and face 2 shaded blue
\[ \frac{1/6}{3/6} = \frac{\textcolor{#1f5fbf}{1}}{3} \]
Check with the shared-6 rule
Why: matches the shrunk-space count
Worked example
Figure (svg): All six faces back in plain ink as S, with 2, 4 and 6 still heavily outlined as B and face 2 shaded blue
\( {\dfrac{\textcolor{#1f5fbf}{1}}{3} = \dfrac{\textcolor{#1f5fbf}{1} \div 6}{3 \div 6}}\;\;\Rightarrow\;\;\allowbreak {\textcolor{#1f5fbf}{1} \div 6 = \textcolor{#1f5fbf}{P(A \text{ AND } B)}}\;\;\Rightarrow\;\;\allowbreak {3 \div 6 = P(B)} \)
\[ P(A \mid B) = \frac{\textcolor{#1f5fbf}{P(A \text{ AND } B)}}{P(B)} \]
Put both chances in place
Why: holds for any equally likely S
Figure (svg): All six faces back in plain ink as S, with 2, 4 and 6 still heavily outlined as B and face 2 shaded blue
The bottom cannot be 0: the rule needs P(B) > 0.
\[ \{\textcolor{#1f5fbf}{4, 6}\}\colon\ \frac{\textcolor{#1f5fbf}{2/6}}{3/6} = \frac{\textcolor{#1f5fbf}{2}}{3} \]
Check: try the event {4, 6} instead
Why: rule and picture must agree
Figure (svg): All six faces as S, with 2, 4 and 6 outlined as B and faces 4 and 6 shaded blue as the test event
Worked example
Figure (svg): The six faces of one fair die with 2 and 3 shaded blue as event A
\[ \textcolor{#1f5fbf}{A} \text{ AND } S = \textcolor{#1f5fbf}{A} \]
Keep A's faces inside S
Why: S holds every face
Figure (svg): The six faces all heavily outlined as the news S, with faces 2 and 3 also shaded blue as A
\[ P(\textcolor{#1f5fbf}{A} \mid S) = P(\textcolor{#1f5fbf}{A} \text{ AND } S) \div P(S) \]
Make S the news in the rule
Why: the rule takes any news
\[ P(\textcolor{#1f5fbf}{A} \mid S) = \textcolor{#1f5fbf}{2/6} \div 6/6 \]
Substitute both chances
Why: the top overlap is A itself
\[ \textcolor{#1f5fbf}{2/6} \div 6/6 = \frac{\textcolor{#1f5fbf}{2}}{6} \]
Cancel the shared 6
Why: compare with the old chance
\[ \textcolor{#1f5fbf}{2} \text{ of } 6 \text{ outlined faces} \]
Check: count blue among outlined
Why: no face ruled out
Worked example
Figure (svg): Tickets 1 to 19 in blocks of seven with rows A and B marked and a blue AND row shading 14, 16 and 18
\( {P(A \text{ AND } B) = \dfrac{\textcolor{#1f5fbf}{3}}{19}},\quad\allowbreak \allowbreak {P(B) = \dfrac{6}{19}} \)
\[ P(A \mid B) = \frac{\textcolor{#1f5fbf}{3/19}}{6/19} \]
Put in the raffle's chances
Why: the news sits in the bottom
Figure (svg): The ticket blocks with a blue AND row at 14, 16, 18 and the columns of B, tickets 14 to 19, boxed
\[ \frac{\textcolor{#1f5fbf}{3/19}}{6/19} = \frac{\textcolor{#1f5fbf}{3}}{6} \]
Cancel the shared 19
Why: leaves counts inside B's six tickets
\[ \textcolor{#1f5fbf}{3} \text{ of } 6 \text{ boxed} \]
Check: count blue in the boxes
Why: counting inside B must agree
Worked example
Figure (svg): The ticket blocks with a blue AND row at 14, 16, 18 and the columns of B, tickets 14 to 19, boxed
\( {P(A \text{ AND } B) = \dfrac{\textcolor{#1f5fbf}{3}}{19}},\quad\allowbreak \allowbreak {P(A) = \dfrac{9}{19}} \)
\[ P(B \mid A) = \frac{\textcolor{#1f5fbf}{3/19}}{9/19} \]
Make A the news instead
Why: tests whether swapping the news matters
Figure (svg): The ticket blocks with a blue AND row at 14, 16, 18 and the columns of A, the even tickets, boxed
\[ \frac{\textcolor{#1f5fbf}{3/19}}{9/19} = \frac{\textcolor{#1f5fbf}{3}}{9} \]
Cancel the 19 again
Why: leaves counts inside A's nine tickets
\[ \textcolor{#1f5fbf}{3} \text{ of } 9 \text{ boxed} \]
Check: tally blue among boxed columns
Why: picture must match the fraction
Trap
Try It 3.2: above 6, given odd?
\[ \text{above 6 and odd: } \{7, 9\} \]
Find the numbers in both
Why: the rule's top count
\[ 2 \div 4 = 0.5 \]
Divide by the 4 above 6
Why: treats 'above 6' as news
\[ 0.5 = P(\text{odd} \mid \text{above } 6) \]
Name what 0.5 measures
Why: fails: the news was 'odd'
Figure (svg): Numbers 1 to 10 with the odd numbers heavily outlined as the news, 7 and 9 also shaded blue, and even numbers greyed out
\[ P(\text{above } 6 \mid \text{odd}) = \frac{\textcolor{#1f5fbf}{2}}{5} \]
Divide by the 5 odd numbers
Why: the given event fills the bottom
\[ 1, 3, 5, 7, 9\colon\ 5;\ \ \textcolor{#1f5fbf}{7, 9}\colon\ \textcolor{#1f5fbf}{2} \]
Check: recount the outlined cells
Why: a recount, not the same division
Prediction
Predict first
Example 3.2: A = even face. News B: the face is under 4.
Is A now more likely, less likely, or unchanged?
Correct: Less likely
Why: Among faces 1, 2 and 3 only 2 is even: one face of three. Before the news, three faces of six were even, which is more. News can raise, lower or leave a chance.
Worked example
Figure (svg): The six faces of one fair die as plain cells 1 to 6
\[ P(A) = \frac{\textcolor{#1f5fbf}{3}}{6} \]
Count even faces over all 6
Why: the old chance, for comparing
Figure (svg): The six faces with the even faces 2, 4 and 6 shaded blue
\[ A \text{ AND } B = \{\textcolor{#1f5fbf}{2}\} \]
Keep even faces under 4
Why: only face 2 passes both tests
Figure (svg): The six faces with 1, 2 and 3 heavily outlined as B, face 2 also blue, and 4, 5, 6 greyed out
\[ P(A \mid B) = \frac{\textcolor{#1f5fbf}{1}}{3} \]
Divide by B's 3 faces
Why: news shrinks the space
\[ \frac{\textcolor{#1f5fbf}{1}}{3} = \frac{\textcolor{#1f5fbf}{2}}{6} \]
Write 1/3 in sixths
Why: compares directly with 3/6
\[ \frac{\textcolor{#1f5fbf}{2}}{6} < \frac{\textcolor{#1f5fbf}{3}}{6} \]
Check: fewer sixths than before
Why: the news lowered A's chance
Faded example
Figure (svg): Try It 3.1's 12 ordered pairs as a 3 by 4 grid: rows are the first number 1 to 3, columns the second number 1 to 4
A: the sum is even. B: first number prime (2 or 3).
Fill in the blanks
pairs in B = 8; pairs in A AND B = 4; P(A | B) = 4/8
Why: B is rows 2 and 3: 2 × 4 = 8 pairs. Even sums there: (2,2), (2,4), (3,1), (3,3), four pairs. So P(A | B) = 4/8 = 0.5.
Worked example
Figure (svg): Try It 3.1's 12 ordered pairs as a 3 by 4 grid: rows are the first number 1 to 3, columns the second number 1 to 4
\[ 2 \times 4 = 8 \text{ pairs in } B \]
Count rows 2 and 3
Why: B's pairs: the new sample space
Figure (svg): The pairs grid with rows 2 and 3 heavily outlined as B and row 1 greyed out
\[ \textcolor{#1f5fbf}{(2,2),\ (2,4),\ (3,1),\ (3,3)} \]
Find even sums inside B
Why: both events hold for these
Figure (svg): The pairs grid with B outlined, and (2,2), (2,4), (3,1), (3,3) shaded blue
\[ P(A \mid B) = \frac{\textcolor{#1f5fbf}{4}}{8} \]
Divide by B's 8 pairs
Why: the shrunk space goes below
\[ \frac{\textcolor{#1f5fbf}{4}}{8} + \frac{4}{8} = 1 \]
Check: add B's 4 odd-sum pairs
Why: B's 8 pairs fill the whole
Pattern
\[ P(A) = \frac{\text{in } A}{\text{in } S},\quad P(A') = 1 - P(A),\quad P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)} \]
Check
Figure (svg): Example 3.3's table: males 43 right-handed, 9 left-handed; females 44 right-handed, 4 left-handed
Example 3.3: F = female, L = left-handed.
Check your understanding
One of these 100 people is picked at random. Find P(L | F).
Answer: A
Why: Given F, only the female row counts: 44 + 4 = 48 people, and 4 of them are left-handed. So P(L | F) = 4/48, about 0.083.
Worked example
Figure (svg): Example 3.3's table: males 43 right-handed, 9 left-handed; females 44 right-handed, 4 left-handed
\[ 44 + \textcolor{#1f5fbf}{4} = 48 \]
Total the female row
Why: the news makes it the space
Figure (svg): Example 3.3's table with a total column (52, 48) and the female row heavily outlined
\[ P(L \mid F) = \frac{\textcolor{#1f5fbf}{4}}{48} \]
Put the 4 left-handers over 48
Why: only the outlined row counts
Figure (svg): The table with the female row outlined and the cell 4 (female, left) shaded blue
\[ \textcolor{#1f5fbf}{4} \div 48 \approx 0.083 \]
Divide 4 by 48
Why: a decimal ranks against other chances
\[ 9 + \textcolor{#1f5fbf}{4} = 13 \]
Total the left-handed column
Why: explains the rival answer 4/13
\[ \frac{44}{48} + \frac{\textcolor{#1f5fbf}{4}}{48} = 1 \]
Check: add the right-handed share
Why: the row fills the whole
Check
A standard deck: 13 ranks in each of 4 suits.
Check your understanding
The drawn card is a jack, queen or king. A classmate says P(king) stays 4/52. Which reply is right?
Answer: A
Why: The news shrinks the space to the 12 face cards (3 ranks in 4 suits), and 4 of them are kings: 4/12 = 1/3, far above 4/52 = 1/13.
Worked example
Figure (svg): The 12 face cards as a grid: rows are the suits spades, hearts, diamonds, clubs; columns are jack, queen, king
\[ 4 \times 3 = 12 \]
Count suits times face ranks
Why: the news leaves 12 cards
Figure (svg): The face-card grid with all 12 cells heavily outlined as the news
\[ P(\text{king} \mid \text{face}) = \frac{\textcolor{#1f5fbf}{4}}{12} \]
Divide the 4 kings by 12
Why: the 12 cards stay equally likely
Figure (svg): The face-card grid outlined, with the king column shaded blue
\[ P(\text{king}) = \frac{4}{52} \]
Count kings in the whole deck
Why: the classmate's chance without news
\[ \frac{\textcolor{#1f5fbf}{4} \div 4}{12 \div 4} = \frac{1}{3},\ \ \frac{4 \div 4}{52 \div 4} = \frac{1}{13} \]
Check: reduce both shares
Why: exposes the gap between shares
Worked example
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells HH, HT, TH, TT
\[ B = \{HH, HT, TH\} \]
Keep cells with a head
Why: the peek rules out TT
Figure (svg): The coin grid with HH, HT, TH heavily outlined as B and TT greyed out
\[ \text{both heads: } \{\textcolor{#1f5fbf}{HH}\} \]
Find HH inside B
Why: the count's top: HH AND B
Figure (svg): The coin grid with B outlined, HH shaded blue and TT greyed out
\[ P(HH \mid B) = \frac{\textcolor{#1f5fbf}{1}}{3} \]
Divide by B's 3 cells
Why: they stay equally likely
\[ P(HH \text{ AND } B) = \frac{\textcolor{#1f5fbf}{1}}{4},\ \ P(B) = \frac{3}{4} \]
Divide both counts by 4
Why: readies the chance rule
\[ \frac{\textcolor{#1f5fbf}{1/4}}{3/4} = \frac{\textcolor{#1f5fbf}{1}}{3} \]
Check: cancel the shared 4
Why: the rule matches counting
Worked example
Figure (svg): A 2 by 2 grid of the two coins' outcomes: rows are the dime's H and T, columns the nickel's H and T, cells HH, HT, TH, TT
\( {B = \{HH, HT, TH\}},\quad\allowbreak \allowbreak {P(HH \mid B) = \dfrac{\textcolor{#1f5fbf}{1}}{3}} \)
\[ D = \{HH, HT\} \]
Keep cells with the dime heads
Why: tests 'the other coin'
Figure (svg): The coin grid with the dime-heads row HH, HT outlined and row TH, TT greyed out
\[ P(HH \mid D) = \frac{\textcolor{#1f5fbf}{1}}{2} \]
Divide by D's 2 cells
Why: the 50-50 the intuition wanted
Figure (svg): The coin grid with the dime-heads row outlined and HH shaded blue
\[ \frac{\textcolor{#1f5fbf}{1}}{3} < \frac{\textcolor{#1f5fbf}{1}}{2} \]
Compare the two kinds of news
Why: which news keeps more cells?
\[ TH\colon\ \text{in } B,\ \text{not in } D \]
Check: find the cell that differs
Why: TH alone explains the gap
Recap
OpenStax Introductory Statistics 2e, §3.1 Terminology §3.1, pp. 168-171 — Examples 3.1–3.3 and Try Its 3.1–3.2 trace back here
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