Theoretical probability as a ratio of counts, probabilities computed with permutations and combinations, odds in favour of and against an event, experimental probability from surveys and trials, and geometric probability computed from lengths and areas.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Define and Use Probability
Objectives
Five outcomes. Counting has become measuring how likely.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 698-703 — the lesson these objectives are drawn from
Warm-up
Lessons 10.1 and 10.2 counted outcomes. Now those counts get compared.
Discussion prompt
A die has six faces. How many show an even number, and what fraction of the faces is that?
Hint: Count the favourable faces, then divide.
Answer:
\[ \text{even faces: } 2, 4, 6 \;\Longrightarrow\; 3 \text{ of } 6 \]
\[ \frac{3}{6} = \frac{1}{2} \]
That fraction is the probability. Everything in this lesson is a ratio of two counts you already know how to find, so the counting work is done and only the interpretation is new.
Concept
When all outcomes are equally likely, the probability of an event is the number of outcomes in it divided by the total number of outcomes. The result always lies between 0 and 1.
theoretical probability — The number of outcomes in an event divided by the total number of outcomes, when all outcomes are equally likely. Experimental probability instead divides the number of trials in which the event occurred by the total number of trials.
\[ P(A) = \frac{\text{outcomes in } A}{\text{total outcomes}} \]
The equally-likely condition matters. A weighted die has six outcomes but they are not equally likely, so counting alone would give the wrong answer.
Figure (svg): A number line from zero to one showing what different probabilities mean
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 698-698
Section
Section 1
Concept
An outcome is one possible result and an event is a collection of outcomes. The probability of an event is the count of its outcomes over the count of all outcomes, provided every outcome is equally likely.
\[ P(\text{even}) = \frac{3}{6} = \frac{1}{2} \]
A probability of 0 means impossible and 1 means certain, with one half the point where an event is as likely to happen as not.
Figure (svg): The six outcomes of a die roll with two events marked as subsets
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 698-698 — Theoretical Probability of an Event
Picture it
Example 1: rolling a 5 and rolling an even number.
Figure (svg): The six outcomes of a die roll with two events marked as subsets
One outcome out of six gives one sixth; three out of six gives one half. The denominator is the same in both because the die is the same.
Worked example
Example 1, both parts.
\[ \text{Rolling a standard die, find } P(\text{a } 5) \text{ and } P(\text{even}). \]
Count the total outcomes
Why: Six faces, all equally likely.
\[ 6 \]
Count the first event
Why: Only one face shows a 5.
\[ 1 \]
Count the second event
Why: Two, 4 and 6 are even.
\[ 3 \]
Divide each and simplify
Why: One sixth, and 3 sixths.
\[ \frac{1}{6}\text{ and } \frac{1}{2} \]
Figure (svg): The six outcomes of a die roll with two events marked as subsets
\[ P = \tfrac{1}{6}; \qquad P = \tfrac{1}{2} \]
Verify: check that both lie between 0 and 1
Why: One sixth is about 0.167 and one half is 0.5, both in range. And rolling an even number should be three times as likely as rolling a 5 specifically, since it covers three faces rather than one — which the two answers confirm.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 698-698
Fill the middle
Example 1b.
Fill in the blanks
P(\text3) = \frac___}___ = \frac______
Why: Three of the six faces are even, so the probability is 3 over 6, which simplifies to one half. The denominator counts every face, favourable or not.
Worked example
Guided Practice 1 and 2.
\[ \text{Choosing an integer from } 1 \text{ to } 20, \text{ find } P(\text{perfect square}) \text{ and } P(\text{factor of } 30). \]
Count the total
Why: Twenty integers, all equally likely.
\[ 20 \]
List the perfect squares
Why: One, 4, 9 and 16.
\[ 4\text{ outcomes} \]
List the factors of 30 up to 20
Why: One, 2, 3, 5, 6, 10 and 15.
\[ 7\text{ outcomes} \]
Divide
Why: Four twentieths and 7 twentieths.
\[ \frac{1}{5}\text{ and } \frac{7}{20} \]
Figure (svg): The solution to Worked example choosing an integer shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{4}{20} = \tfrac{1}{5}; \qquad \tfrac{7}{20} \]
Verify: check the factor list
Why: Thirty's factors are 1, 2, 3, 5, 6, 10, 15 and 30, and only the first seven are at most 20. Missing 15 or wrongly including 30 are the two easy slips, so listing every factor of 30 first and then striking out the ones above 20 is the safer order.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 699-699
Trap
\[ \text{roll two dice; the total is } 2, 3, \dots, 12 \]
Take each total as one outcome
Why: Eleven totals are listed, so each is given probability one eleventh.
\[ P(\text{total} = 7) = \tfrac{1}{11} \quad \text{(wrong)} \]
A total of 7 happens in six ways and a total of 2 in only one, so the eleven totals are not equally likely.
\[ 36 \text{ equally likely pairs}; \; 6 \text{ give a total of } 7 \]
Count the equally likely outcomes, which are the pairs
Why: The formula needs a list in which every entry is as likely as every other.
\[ P(\text{total} = 7) = \tfrac{6}{36} = \tfrac{1}{6} \]
Choosing what counts as an outcome is the real decision. Getting it wrong makes a perfectly correct division give a wrong answer.
Sorting
Compare each probability with one half.
Sort into buckets
Sort each event for one roll of a fair die.
The fourth is certain, with probability 1, since every face is under 7. Both extremes are legitimate probabilities rather than special cases.
Matching
Choosing an integer from 1 to 20.
Match the pairs
Why: The last has probability zero because no outcome qualifies. Notice the denominators are all 20: the total does not depend on which event is being asked about.
Prediction
Commit before reasoning.
Predict first
Why does the counting formula require every outcome to be equally likely?
Correct: Because otherwise a count says nothing about how often each outcome happens.
\[ 6 \text{ faces} \neq 6 \text{ equally likely outcomes, if weighted} \]
Why: A weighted die still has six faces, so the count is unchanged, but a face weighted to appear half the time deserves probability one half rather than one sixth. Counting works only when each item in the list carries the same weight. This is why the two-dice trap arises: the eleven totals are a valid list of results but not a valid list of equally likely outcomes.
Section
Section 2
Concept
When the outcomes are too many to list, count them with the formulas of the previous two lessons. Order matters for some events and not for others, so both kinds of count appear.
\[ P = \frac{\;_4C_2}{\;_7C_2} = \frac{6}{21} = \frac{2}{7} \]
Whichever formula is used, it must be used in both the numerator and the denominator, so that the two counts describe the same kind of outcome.
Figure (svg): Two probabilities computed from counting formulas rather than by listing
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 699-699 — Use permutations or combinations
Picture it
Example 2: seven musicians in a random order.
Figure (svg): Two probabilities computed from counting formulas rather than by listing
The alphabetical event needs orderings, so 7 factorial; the pair of friends does not, so 7 choose 2. Mixing the two would compare unlike things.
Worked example
Example 2, both parts.
\[ \text{With } 7 \text{ musicians in random order, find } P(\text{alphabetical}) \text{ and } P(\text{first two are your } 4 \text{ friends}). \]
Alphabetical: count all orders
Why: Seven factorial.
\[ 5040 \]
Alphabetical: count the favourable
Why: Exactly one order is alphabetical.
\[ \frac{1}{5040} \]
Friends: count all pairs
Why: Order of the first two does not matter here.
\[ 7 C 2 = 21 \]
Friends: count the favourable pairs
Why: Two of your 4 friends.
\[ 4 C 2 = 6,\text{ so } \frac{6}{21} \]
Figure (svg): Two probabilities computed from counting formulas rather than by listing
\[ \tfrac{1}{5040}; \qquad \tfrac{6}{21} = \tfrac{2}{7} \]
Verify: check the second differently
Why: The chance the first performer is a friend is 4 over 7, and then 3 of the remaining 6 are friends, giving 4 over 7 times 3 over 6, which is 2 over 7. The combination route and this step-by-step route agree, which confirms that treating the pair as unordered was legitimate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 699-699
Fill the middle
Example 2b.
Fill in the blanks
P = \frac6___ = \frac___}___
Why: Four choose 2 is 6, the number of ways to pick two of your four friends. The denominator counts every possible pair of the seven performers.
Worked example
Guided Practice 3.
\[ \text{Repeat with } 9 \text{ musicians.} \]
Alphabetical: count all orders
Why: Nine factorial.
\[ 362, 880 \]
Alphabetical: the probability
Why: Still exactly one favourable order.
\[ \text{about } 0.0000028 \]
Friends: count all pairs
Why: Nine choose 2.
\[ 36 \]
Friends: the probability
Why: Still 6 favourable pairs.
\[ \frac{6}{36} = \frac{1}{6} \]
Figure (svg): The solution to Worked example nine musicians instead shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{362{,}880}; \qquad \tfrac{1}{6} \approx 0.167 \]
Verify: compare the two changes
Why: Adding two musicians divided the alphabetical probability by 72, since 9 factorial is 72 times 7 factorial. The friends probability fell only from 0.286 to 0.167, a much gentler drop, because the favourable count stayed at 6 while the total rose from 21 to 36. Factorials in a denominator collapse a probability far faster than combinations do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 699-699
Error analysis
A student computes the probability that the first two of seven performers are friends.
Annotate
On: \( P = \frac{\;_4C_2}{\;_7P_2} = \frac{6}{42} = \frac{1}{7} \)
Either count both ordered or count both unordered. Using permutations in both would give 12 over 42, which is also 2 over 7 — the same answer, as it must be.
Sorting
Would rearranging change the event?
Sort into buckets
Sort each probability question.
Either kind of count gives the right probability so long as it is used in both the numerator and the denominator — the choice is about convenience, not correctness.
Comparison
Fill the blanks. Same events, larger group.
Comparison matrix
| Event | 7 musicians | 9 musicians |
|---|---|---|
| Alphabetical order | 1/5040 | 1/362,880 |
| First two are friends | 6/21 = 2/7 | 6/36 = 1/6 |
| Which fell faster | the alphabetical one, by a factor of 72 | the friends one, by under half |
| Why | a factorial denominator | a combination denominator, which grows more slowly |
Both denominators grew, but factorials grow so much faster that the first probability became negligible while the second stayed easily observable.
Prediction
Commit before reasoning.
Predict first
Example 2b used combinations. What if permutations were used in both places instead?
Correct: The same answer, since the r factorial cancels between numerator and denominator.
\[ \frac{r!\cdot \;_4C_2}{r!\cdot \;_7C_2} = \frac{\;_4C_2}{\;_7C_2} \]
Why: Four P 2 is 12 and 7 P 2 is 42, giving 12 over 42, which is 2 over 7 — the same as 6 over 21. Each count is r factorial times its combination, and that factor appears in both parts of the fraction, so it cancels. The rule is therefore not which formula to use but to use the SAME one in both places.
Section
Section 3
Concept
The odds in favour of an event compare the number of outcomes in it with the number not in it. The odds against reverse that ratio. Neither is the probability, which compares the event with the total.
\[ \text{odds in favour} = \frac{\text{in } A}{\text{not in } A} \]
Odds are usually written with a colon, as one to twelve, and the two numbers add to the total rather than one of them being it.
Figure (svg): Odds compared with probability for the same event, showing the different denominators
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 699-700 — Odds in Favor of or Odds Against an Event
Picture it
Example 3: drawing a 10 from a standard deck.
Figure (svg): Odds compared with probability for the same event, showing the different denominators
The probability is 4 over 52 and the odds in favour are 4 over 48. The book prints a caution beside this example for exactly that reason.
Worked example
Example 3, both parts.
\[ \text{From a } 52 \text{-card deck, find the odds in favour of drawing a } 10 \text{ and the odds against drawing a club.} \]
Count tens and non-tens
Why: Four tens, 48 others.
\[ 4\text{ to } 48 \]
Simplify
Why: Divide both by 4.
\[ 1: 12 \]
Count non-clubs and clubs
Why: Thirty-nine others, 13 clubs.
\[ 39\text{ to } 13 \]
Simplify
Why: Divide both by 13.
\[ 3: 1 \]
Figure (svg): Odds compared with probability for the same event, showing the different denominators
\[ 1:12; \qquad 3:1 \]
Verify: compare each with the probability
Why: The probability of a 10 is 4 over 52, or 1 over 13, while the odds are 1 to 12 — close but not equal. The probability of a club is 13 over 52, or one quarter, while the odds against are 3 to 1. In each case the two numbers of the odds add to the total, whereas the probability's denominator IS the total.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 700-700
Fill the middle
Example 3a.
Fill in the blanks
\text48 10 = \frac______} = 1:12
Why: Forty-eight cards are not tens, so the odds are 4 to 48, which simplifies to 1 to 12. The denominator of odds excludes the favourable outcomes.
Worked example
Guided Practice 4 and 5.
\[ \text{Find the odds in favour of drawing a heart and the odds against drawing a queen.} \]
Count hearts and non-hearts
Why: Thirteen hearts, 39 others.
\[ 13\text{ to } 39 \]
Simplify
Why: Divide both by 13.
\[ 1: 3 \]
Count non-queens and queens
Why: Forty-eight others, 4 queens.
\[ 48\text{ to } 4 \]
Simplify
Why: Divide both by 4.
\[ 12: 1 \]
Figure (svg): The solution to Worked example two more sets of odds shown as a ladder of expressions, one row per algebraic move
\[ 1:3; \qquad 12:1 \]
Verify: convert one back to a probability
Why: Odds of 1 to 3 in favour mean 1 favourable outcome for every 3 unfavourable, so 1 in every 4 draws, which is a probability of one quarter — matching 13 over 52. Converting between the two is a good check: odds of a to b correspond to a probability of a over a plus b.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 700-700
Trap
\[ \text{odds in favour of a } 10 = 1:12 \]
Report the probability as one twelfth
Why: The odds ratio is read as a probability.
\[ P = \tfrac{1}{12} \quad \text{(wrong)} \]
The probability is 4 over 52, which is 1 over 13. The odds compare tens with non-tens; the probability compares tens with all cards.
\[ P = \tfrac{4}{52} = \tfrac{1}{13}; \qquad \text{odds} = \tfrac{4}{48} = 1:12 \]
Divide by the total for probability and by the rest for odds
Why: The two denominators differ by exactly the favourable count.
\[ \text{odds } a:b \;\Longleftrightarrow\; P = \tfrac{a}{a+b} \]
The two are close when the event is unlikely and quite different when it is common — odds of 3 to 1 against a club correspond to a probability of one quarter, not one third.
Comparison
Fill the blanks. Different denominators.
Comparison matrix
| Event | Probability | Odds in favour |
|---|---|---|
| Drawing a 10 | 4/52 = 1/13 | 4/48 = 1 : 12 |
| Drawing a heart | 13/52 = 1/4 | 13/39 = 1 : 3 |
| Denominator counts | all outcomes | only the unfavourable ones |
| Converting | P = a/(a + b) | odds = P/(1 - P) |
The last row is the pair of conversions, and either one turns a stated odds into a probability or back again in a single step.
Matching
Favourable against unfavourable.
Match the pairs
Why: Notice the first and last are reverses of each other, since tens and queens are equally common — the odds in favour of one are the reverse of the odds against the other. Reversing the ratio switches between in favour and against.
Prediction
Commit before reasoning.
Predict first
For which events do the odds in favour and the probability nearly agree?
Correct: For rare events, since the unfavourable count is then close to the total.
\[ P = \frac{a}{a+b}; \quad a \text{ small} \;\Longrightarrow\; a+b \approx b \]
Why: The odds of a 10 are 1 in 12 and the probability is 1 in 13 — very close, because 48 and 52 differ by little. For a common event the gap is large: odds of 1 to 1 correspond to a probability of one half, not 1. The two denominators differ by exactly the favourable count, so the rarer the event, the smaller that difference matters.
Section
Section 4
Concept
When the outcomes are not equally likely, or the structure is unknown, probability can be estimated from data: the number of trials in which the event occurred, divided by the total number of trials.
\[ P(A) = \frac{\text{trials where } A \text{ occurs}}{\text{total trials}} \]
The formula looks identical to the theoretical one, but the two numbers come from observation rather than from counting possibilities.
Figure (svg): Two columns comparing theoretical probability with experimental probability
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 700-700 — Experimental Probability of an Event
Picture it
Computed in advance, or measured afterwards.
Figure (svg): Two columns comparing theoretical probability with experimental probability
Theoretical probability is fixed by the structure; experimental probability varies from one sample to the next and settles down only as the sample grows.
Worked example
Example 4.
\[ \text{Of } 463, 1085, 879, 551, 300 \text{ and } 238 \text{ adults in six age bands, find } P(\text{at least } 40). \]
Total the survey
Why: Add all six band counts.
\[ 3516 \]
Identify the wanted bands
Why: Forty to 49, 50 to 59 and 60 to 69.
\[ 551, 300, 238 \]
Add those
Why: Five hundred fifty-one plus 300 plus 238.
\[ 1089 \]
Divide
Why: One thousand eighty-nine over 3516.
\[ \text{about } 0.310 \]
Figure (svg): The solution to Worked example a survey probability shown as a ladder of expressions, one row per algebraic move
\[ \frac{1089}{3516} \approx 0.310 \]
Verify: check the totals add up
Why: The three unwanted bands hold 463 plus 1085 plus 879, which is 2427, and 2427 plus 1089 is 3516 — the whole survey. Every respondent falls into exactly one band, so the two groups must account for everyone, and checking that catches a miscounted band.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 700-700
Fill the middle
Example 4.
Fill in the blanks
463+1085+879+551+300+238 = 3516
Why: The six bands total 3516 respondents, which is the denominator for every probability from this survey. Every respondent is counted exactly once.
Worked example
Guided Practice 6.
\[ \text{Find } P(\text{at most } 39) \text{ and } P(\text{at least } 30). \]
At most 39: identify the bands
Why: Under 20, 20 to 29 and 30 to 39.
\[ 463, 1085, 879 \]
At most 39: add and divide
Why: Two thousand four hundred twenty-seven over 3516.
\[ \text{about } 0.690 \]
At least 30: identify the bands
Why: Everything except the two youngest.
\[ 879 + 551 + 300 + 238 \]
At least 30: add and divide
Why: Three thousand fifty-three over 3516.
\[ \text{about } 0.868 \]
Figure (svg): The solution to Worked example two more from the same data shown as a ladder of expressions, one row per algebraic move
\[ \frac{2427}{3516} \approx 0.690; \qquad \frac{3053}{3516} \approx 0.868 \]
Verify: check the first against Example 4
Why: At most 39 and at least 40 are opposites, so their probabilities should add to 1. They give 0.690 plus 0.310, which is exactly 1. That complement check is the fastest available and it uses no extra data.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 700-700
Error analysis
A student computes the probability that a surveyed adult prefers to be at least 40.
Annotate
On: \( P = \frac{1089}{1000} = 1.089 \)
Any probability outside the range from 0 to 1 signals a wrong denominator, and checking the range costs nothing.
Sorting
Counted from structure, or measured from data?
Sort into buckets
Sort each probability.
Most real probabilities are experimental. Dice and cards are unusual precisely because their structure guarantees equally likely outcomes.
Prediction
Commit before reasoning.
Predict first
The survey of 3516 adults gives 0.310. What would a survey of 35,160 comparable adults be expected to give?
Correct: Something close to 0.310, and more reliably so than a small survey would.
\[ \text{larger sample} \;\Longrightarrow\; \text{less variation, not a different target} \]
Why: Experimental probability estimates an underlying tendency, and larger samples estimate it more precisely. A survey of ten adults might easily give 0.2 or 0.4 by chance; one of 35,160 would rarely stray far from the true value. This is why sample size is reported alongside survey results, and why a theoretical probability, which is exact, needs no sample size at all.
Matching
Add the bands, then divide by 3516.
Match the pairs
Why: The first two are complements and add to 1, and the last two also happen to add to 1 since under 20 and at least 30 do not overlap and... in fact they leave out the twenties, so they sum to 1 minus the 20 to 29 share. Checking which pairs are genuine complements is worth a moment.
Section
Section 5
Concept
When outcomes vary continuously, a count is impossible. Instead the probability is the ratio of the favourable measure to the total measure — a length, an area or a volume.
\[ P = \frac{\text{favourable area}}{\text{total area}} \]
The assumption that replaces equal likelihood is uniformity: every point is as likely as every other, so equal areas carry equal probability.
Figure (svg): A square dartboard with concentric scoring rings and two regions compared by area
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 701-701 — Geometric probability
Picture it
Example 5: an 18 inch square with rings at radii 3, 6 and 9.
Figure (svg): A square dartboard with concentric scoring rings and two regions compared by area
Scoring 5 is the most likely of the three shown, then 0, then 10 — even though the 10 ring sits in the middle where you are aiming.
Worked example
Example 5.
\[ \text{On an } 18 \text{ inch square board with a } 3 \text{ inch inner circle and a } 9 \text{ inch outer circle, compare } P(10) \text{ and } P(0). \]
Find the board's area
Why: Eighteen squared.
\[ 324 \]
Find the inner circle's area
Why: Pi times 3 squared.
\[ 9 \pi \]
Compute the first probability
Why: Nine pi over 324.
\[ \frac{\pi}{36},\text{ about } 0.087 \]
Compute the second
Why: The square minus the largest circle, over the square.
\[ \frac{324 - 81 \pi}{324},\text{ about } 0.215 \]
Figure (svg): A square dartboard with concentric scoring rings and two regions compared by area
\[ 0.0873 < 0.215 \;\Longrightarrow\; 0 \text{ points more likely} \]
Verify: check the corners argument
Why: The largest circle has area 81 pi, about 254, out of 324 — so the four corners left over hold about 70 square inches, more than a fifth of the board. Corners are easy to underestimate by eye, which is exactly why the area computation is needed rather than an impression.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 701-701
Fill the middle
Guided Practice 7.
Fill in the blanks
\pi(6^2)-\pi(3^2) = 27\pi
Why: Thirty-six minus 9 is 27, so the ring's area is 27 pi. Subtracting the inner circle from the outer one is how every ring's area is found.
Worked example
Guided Practice 7.
\[ \text{Compare } P(5 \text{ points}) \text{ with } P(0 \text{ points}). \]
Identify the 5-point region
Why: Between the circles of radius 3 and 6.
Find its area
Why: Pi times 36 minus pi times 9.
\[ 27 \pi \]
Compute the probability
Why: Twenty-seven pi over 324.
\[ \frac{\pi}{12},\text{ about } 0.262 \]
Compare with zero points
Why: Point two six two against 0.215.
\[ 5\text{ points is more likely} \]
Figure (svg): A square dartboard with concentric scoring rings and two regions compared by area
\[ \frac{27\pi}{324} = \frac{\pi}{12} \approx 0.262 \]
Verify: check why the ring beats the bullseye so easily
Why: The 5-point ring is only 3 inches wide, the same as the bullseye's radius, yet its area is three times as large — because area grows with the square of the radius, so an outer ring covers far more ground than an inner one of the same width. That is the whole reason the middle rings dominate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 701-701
Trap
\[ \text{bullseye radius } 3; \; \text{ring width } 3 \]
Conclude the two regions are equally likely
Why: Both measure 3 inches, so they are treated as equal.
\[ P(10) = P(5) \quad \text{(wrong)} \]
The ring covers 27 pi and the bullseye only 9 pi. Equal widths do not mean equal areas once you move away from the centre.
\[ P(10) = \tfrac{9\pi}{324}; \quad P(5) = \tfrac{27\pi}{324} \]
Compute each area properly
Why: An annulus is the difference of two circle areas.
\[ \pi(6^2-3^2) = 27\pi = 3 \times 9\pi \]
This is why dartboards and archery targets reward the centre so heavily: hitting it is genuinely much harder than hitting a ring of the same width.
Ranking
Smallest chance first.
Put in order
Why: The areas are about 28, 70, 85, 141 and 324. Each ring is larger than the one inside it even though all three are 3 inches wide, because area grows with the square of the distance from the centre — which is also why the outermost ring alone covers more than the two inner regions together.
Sorting
Length, area or volume.
Sort into buckets
Sort each geometric probability.
Time behaves like length here, which is why waiting-time problems use the same reasoning as points on a rod.
Prediction
Commit before reasoning.
Predict first
The bullseye and the 5-point ring are both 3 inches wide. Why is the ring three times as likely?
Correct: Because area grows with the square of the radius, so an outer band covers far more ground.
\[ \pi(9-0) : \pi(36-9) : \pi(81-36) = 9 : 27 : 45 \]
Why: The bullseye covers pi times 9 and the ring covers pi times 36 minus pi times 9, which is pi times 27 — exactly three times as much. The next ring out covers 45 pi, five times the bullseye. Equal widths give steadily growing areas as you move outward, which is the same squaring relationship that made the surface-area-to-volume ratio fall in Lesson 8.4.
Comparison
Fill the blanks. All four are ratios.
Comparison matrix
| Kind | Numerator | Denominator |
|---|---|---|
| Theoretical | outcomes in the event | all outcomes |
| Odds in favour | outcomes in the event | outcomes NOT in the event |
| Experimental | trials where it occurred | total trials |
| Geometric | favourable length, area or volume | total length, area or volume |
Only the odds row has a denominator that is not the whole, which is exactly why it is not a probability and can exceed 1.
Pattern
Decide which kind, then count or measure.
Every probability lies between 0 and 1. A value outside that range means the wrong denominator was used.
OpenStax Algebra and Trigonometry 2e, §13.7 Probability §13.7
Check
Theoretical probability. Count both parts.
Check your understanding
Choosing an integer from 1 to 20, what is the probability of a perfect square?
Answer: A
Why: The perfect squares are 1, 4, 9 and 16, so the probability is 4 over 20.
Check
Odds are not probability.
Check your understanding
What are the odds in favour of drawing a 10 from a standard deck?
Answer: A
Why: There are 4 tens and 48 non-tens, and 4 to 48 simplifies to 1 to 12.
Check
Geometric probability. Compare areas.
Check your understanding
On an 18 inch square board, what is the probability of hitting a circle of radius 3 at the centre?
Answer: A
Why: The areas are 9 pi and 324, giving pi over 36.
Real world
A medical test for a condition affecting 1 person in 1000 is correct 99 percent of the time, both for people who have the condition and for those who do not. You test positive.
Discussion prompt
Out of 100,000 people, count how many test positive and how many of those actually have the condition, then give the probability that a positive result is correct.
Hint: Work with counts rather than percentages.
Answer:
\[ \text{have it: } 100 \text{ people; } 99 \text{ test positive} \]
\[ \text{do not: } 99{,}900 \text{ people; } 1\% = 999 \text{ test positive} \]
\[ P(\text{have it} \mid \text{positive}) = \frac{99}{99+999} = \frac{99}{1098} \approx 0.090 \]
Only about 9 percent of positive results are correct, even though the test is 99 percent accurate. The false positives outnumber the true ones by more than ten to one.
The reason is that the condition is rare: there are a thousand times more healthy people to generate false positives than sick people to generate true ones. This is exactly the theoretical-probability formula — favourable outcomes over total outcomes — applied to a carefully counted population, and it is why screening programmes for rare conditions always retest before treating. The intuition that a 99 percent accurate test gives a 99 percent reliable positive is one of the most consistently wrong intuitions in all of probability, and counting is what corrects it.
Commit first
Answer, then rate your confidence honestly.
Predict first
The odds in favour of drawing a 10 are 1 to 12. Is the probability one twelfth?
Correct: No — the probability is 4 over 52, which is one thirteenth.
\[ \text{odds } 1:12 \;\Longrightarrow\; P = \frac{1}{1+12} = \frac{1}{13} \]
Why: Odds compare the favourable outcomes with the unfavourable ones, 4 against 48. Probability compares the favourable outcomes with everything, 4 against 52. The two denominators differ by exactly the 4 favourable cards. Odds of a to b always correspond to a probability of a over a plus b, so 1 to 12 gives 1 over 13 — close, but not equal. For a common event the gap is much larger: odds of 1 to 1 mean a probability of one half, not 1.
Explain it
They think probability is just guessing how likely something feels.
Discussion prompt
In four sentences or fewer, explain how to compute a probability for a fair die.
Hint: Count twice.
Answer:
List everything that could happen — for a die, the six faces — and then count how many of those are the thing you want.
The probability is that count divided by the total. Rolling an even number covers three of the six faces, so the probability is 3 over 6, which is one half. It only works if every outcome is equally likely, which is what makes a fair die a good example.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For outcomes, ask whether each item on your list is genuinely as likely as every other. For counting, decide once whether order matters and use that decision in both places. For odds, remember that its denominator excludes the favourable outcomes. For geometric problems, compute both measures explicitly rather than comparing widths by eye.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a probability page. Top: draw the 0-to-1 number line with the five landmarks labelled, and beside it write the theoretical probability formula with a one-line note on the equally-likely condition. Middle left: work both parts of the musician problem, marking which uses a permutation and which a combination, and show that using permutations in both gives the same answer. Middle right: build a small table of four events with their probabilities and their odds side by side, and write the conversion between the two. Bottom left: compute one experimental probability from the survey and check it against its complement. Bottom right: draw the dartboard to scale, compute all four region areas, and rank them, then write one sentence explaining why equal widths give unequal areas.
If any of your probabilities came out above 1, the denominator is wrong: it must count every outcome, not just the unfavourable ones.
Recap
Five things, all of them ratios.
| If you see | Then |
|---|---|
| Equally likely outcomes | Count the event over the total |
| Too many to list | Count with permutations or combinations, the same kind in both |
| The word odds | Divide by the unfavourable count, not the total |
| Survey or trial data | Occurrences over total trials |
| A continuous region | A ratio of lengths, areas or volumes |
| A result outside 0 to 1 | The denominator is wrong |
Lesson 10.4 handles events that overlap, where simply adding the two counts would double-count the outcomes they share.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.3 Define and Use Probability §10.3, pp. 698-703 — everything on these slides traces back here
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