Defines probability as a ratio of counts under the equally-likely assumption, then develops the rules for combining events: subtraction for overlapping unions, multiplication for independent events with the independence condition made explicit, and the complement rule for at-least-one questions.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.7 Probability, pp. 1366-1376
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1366-1376 — the pages these objectives are drawn from
Warm-up
The answer is a counting statement plus an assumption.
Discussion prompt
Rolling a three has probability one sixth. Where do the one and the six come from?
Hint: And what has to be true for the ratio to mean anything?
Answer:
The six counts all the outcomes and the one counts the favourable ones. Both are counts, so §11.5 supplies them.
But the ratio is only a probability if every outcome is equally likely. A loaded die still has six outcomes.
So probability here is counting plus an assumption, and the assumption is the part most often left unstated. Every formula in this section presupposes it.
Concept
When every outcome is equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes.
probability — the ratio of favourable outcomes to total outcomes, when all outcomes are equally likely
\[ P(A)=\frac{\text{favourable}}{\text{total}} \]
Because both numbers are counts, the whole of the previous sections applies directly. The genuinely new content is the rules for combining events, which is where the section's difficulties are.
Figure (svg): A card giving probability as a ratio of favourable outcomes to total outcomes, with the equally-likely assumption flagged
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1366-1369
Section
Section 1
Concept
Computing a probability means counting the favourable outcomes and the total outcomes, and requires that every outcome be equally likely.
The equally-likely assumption is what makes counting sufficient. Without it the outcomes have to be weighted individually, and the ratio of counts says nothing — which is the difference between a fair die and a loaded one.
Figure (svg): A card giving probability as a ratio of favourable outcomes to total outcomes, with the equally-likely assumption flagged
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1366-1370
Picture it
The red line is the assumption.
Figure (svg): A card giving probability as a ratio of favourable outcomes to total outcomes, with the equally-likely assumption flagged
The bottom line records what this section inherits. Both counts come from §11.5, so the counting difficulties carry over along with the counting techniques.
Worked example
Two counts.
\[ \text{What is the probability of drawing a heart from a standard deck?} \]
Count the total outcomes
Why: The whole deck.
\[ 52 \]
Count the favourable ones
Why: One suit.
\[ 13 \]
Check the assumption
Why: A shuffled deck.
Form the ratio
Why: And simplify.
\[ \frac{13}{52} = \frac{1}{4} \]
Figure (svg): A card giving probability as a ratio of favourable outcomes to total outcomes, with the equally-likely assumption flagged
\[ \tfrac{1}{4} \]
Verify: check against the structure
Why: There are four suits of equal size, so each should have probability one quarter — and they do. That the four suit probabilities sum to one is a further consistency check.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1367-1369
Prediction
Any probability computed as a ratio of counts.
Predict first
What values can it take?
Correct: Between zero and one inclusive.
Why: The favourable outcomes are a subset of the total, so the numerator never exceeds the denominator and both are non-negative. A computed probability outside that range signals a counting error.
Worked example
The counts are selection counts.
\[ \text{From } 10 \text{ people, } 4 \text{ are chosen at random. What is the probability a particular person is chosen?} \]
Count the total selections
Why: Four from ten.
\[ C(10, 4) = 210 \]
Count the favourable ones
Why: That person plus three from nine.
\[ C(9, 3) = 84 \]
Form the ratio
Why: Favourable over total.
\[ \frac{84}{210} \]
Simplify
Why: Divide through.
\[ \frac{2}{5} \]
Figure (svg): The solution to Worked example counting with combinations shown as a ladder of expressions, one row per legal move
\[ \tfrac{2}{5} \]
Verify: check against a simpler argument
Why: Four of the ten are chosen, so any particular person has a four in ten chance — which is two fifths, matching. Two independent routes agreeing confirms the combination counting.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1369-1370
Trap
\[ \text{two dice sum to }2,\;3,\;\ldots,\;12: \text{ eleven outcomes, so }P(7)=\tfrac{1}{11} \]
Count the possible totals as outcomes
Why: The sums are treated as equally likely.
A total of seven arises in six ways and a total of two in only one.
Count the equally likely outcomes, which for two dice are the thirty-six ordered pairs.
Six of those give a total of seven, so the probability is six thirty-sixths — one sixth, not one eleventh.
The assumption has to be checked against the outcomes chosen, not assumed to hold for whatever list is convenient.
Faded example
Thirteen hearts in fifty-two cards.
Fill in the blanks
P=\frac524}=\frac______}
Why: The favourable count over the total count gives the probability, provided every card is equally likely to be drawn. Simplifying makes the answer easier to check against intuition.
Sorting
The assumption has to be checked.
Sort into buckets
Sort each set of outcomes.
Step zero
You are asked for a probability.
Discussion prompt
What do you establish before counting?
Hint: What has to be true?
Answer:
What the equally likely outcomes are. The ratio is a probability only for a set of outcomes with no one favoured.
For two dice that means the thirty-six ordered pairs, not the eleven possible totals — those are not equally likely.
Choosing the wrong outcome set gives a plausible-looking wrong answer with no warning. Naming the outcome set explicitly is what makes the counting that follows meaningful.
Section
Section 2
Concept
The probability that either of two events happens is the sum of their probabilities minus the probability of both, because outcomes in both would otherwise be counted twice.
Events that cannot both happen are called mutually exclusive, and for them the subtraction is unnecessary. Checking whether an outcome could satisfy both events is what decides whether the correction is needed.
Figure (svg): Two overlapping circles with the shared region marked, showing why the overlap is subtracted when combining probabilities
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1370-1373
Picture it
The shared region belongs to both circles.
Figure (svg): Two overlapping circles with the shared region marked, showing why the overlap is subtracted when combining probabilities
Adding the two circles counts the middle region twice, so one copy has to come back out. The subtraction is a counting correction rather than an adjustment factor.
Worked example
The overlap must be removed.
\[ \text{A card is drawn. What is the probability it is a heart or a face card?} \]
Count the hearts
Why: One suit.
\[ 13 \]
Count the face cards
Why: Three per suit.
\[ 12 \]
Count the overlap
Why: Face cards that are hearts.
\[ 3 \]
Combine
Why: Add and subtract the overlap.
\[ \frac{13 + 12 - 3}{52} \]
Figure (svg): Two overlapping circles with the shared region marked, showing why the overlap is subtracted when combining probabilities
\[ \tfrac{22}{52}=\tfrac{11}{26} \]
Verify: check by listing the count
Why: Thirteen hearts plus nine non-heart face cards is twenty-two cards, which is the same numerator reached differently. Counting the union directly and by the rule agreeing confirms the overlap was handled correctly.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1371-1372
Prediction
Two events that cannot both happen.
Predict first
What does the union rule become?
Correct: Simple addition, since the overlap is zero.
Why: Mutually exclusive events share no outcomes, so nothing is double counted and the subtraction term vanishes. The general rule still applies; it just reduces.
Worked example
No overlap to subtract.
\[ \text{What is the probability a die shows a } 2 \text{ or a } 5? \]
Check for overlap
Why: A roll cannot be both.
Add the probabilities
Why: One sixth each.
\[ \frac{1}{6} + \frac{1}{6} \]
Note no subtraction
Why: The overlap is empty.
Simplify
Why: The total.
\[ \frac{1}{3} \]
Figure (svg): The solution to Worked example mutually exclusive events shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{3} \]
Verify: check by counting outcomes
Why: Two of the six faces are favourable, giving two sixths — the same answer. The subtraction term was zero, which is what mutually exclusive means, so the rule reduced to simple addition.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1372-1373
Error analysis
A student combines two overlapping events.
Annotate
On: \( P(\text{heart or face})=\tfrac{13}{52}+\tfrac{12}{52}=\tfrac{25}{52} \)
The overlap is counted twice whenever two events share outcomes, which is most of the time. Checking whether any outcome satisfies both events, before adding, is what identifies the need for the correction.
Faded example
Thirteen hearts, twelve face cards, three shared.
Fill in the blanks
\frac3}}22=\frac___}___
Why: The three cards that are both hearts and face cards appear in each of the first two counts, so one copy is removed. The result matches a direct listing of the favourable cards.
Sorting
Can one outcome satisfy both?
Sort into buckets
Sort each pair.
Explain it to yourself
The union rule removes a term.
Discussion prompt
Explain why, and when it is not needed.
Hint: Where do the shared outcomes appear?
Answer:
Outcomes satisfying both events appear in both counts, so adding the two probabilities counts them twice.
Subtracting the probability of both removes exactly one of those copies, leaving each outcome counted once.
When no outcome satisfies both — mutually exclusive events — that term is zero and the rule reduces to addition. A good explanation notes that the general rule always applies, and the exclusive case is just where the correction happens to vanish.
Section
Section 3
Concept
The probability that both of two events happen is the product of their probabilities, but only when the first does not change the second's likelihood.
The distinction is invisible in the arithmetic, which is what makes it dangerous. Multiplying two probabilities produces a number whether or not they are independent, and nothing about the result indicates which case applied.
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1373-1375
Picture it
The last row on each side is the practical test.
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
Replacement is the usual signal. Putting an item back restores the original situation, and not putting it back changes both counts for the second draw.
Worked example
Two separate rolls.
\[ \text{What is the probability of two sixes in two rolls of a die?} \]
Check independence
Why: The first roll does not affect the second.
Find each probability
Why: One sixth each.
\[ \frac{1}{6}\text{ and } \frac{1}{6} \]
Multiply
Why: Since they are independent.
\[ \frac{1}{36} \]
Interpret
Why: One outcome of thirty-six.
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
\[ \tfrac{1}{36} \]
Verify: check against the outcome count
Why: Two rolls have thirty-six equally likely ordered pairs, and exactly one is a double six — giving one thirty-sixth directly. The multiplication and the counting agree, which is what independence guarantees.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1373-1374
Sorting
Does the first affect the second?
Sort into buckets
Sort each pair of events.
Worked example
The second probability changes.
\[ \text{Two cards are drawn without replacement. What is the probability both are hearts?} \]
Find the first probability
Why: Thirteen of fifty-two.
\[ \frac{1}{4} \]
Update for the second
Why: One heart gone, one card gone.
\[ \frac{12}{51} \]
Check independence
Why: The first draw changed the second.
Multiply the updated values
Why: Using the corrected second.
\[ 1 / 4 \times 12 / 51 \]
Figure (svg): The solution to Worked example dependent events shown as a ladder of expressions, one row per legal move
\[ \tfrac{1}{17} \]
Verify: compare with the independent computation
Why: Treating them as independent would give one sixteenth, which is slightly too large — because removing a heart makes the second heart less likely. The two answers are close enough that the error would not look obviously wrong, which is why the check matters.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1374-1375
Trap
\[ P(\text{two hearts})=\tfrac{1}{4}\times\tfrac{1}{4}=\tfrac{1}{16} \]
Multiply the two probabilities directly
Why: Independence is assumed rather than checked.
The first draw removed a heart, so the second probability is lower.
Check whether the first event changes the second. Drawing without replacement always does.
Update the second probability to reflect what the first draw removed, then multiply.
The two answers are close, so the error does not look obviously wrong — which is exactly why the check has to be explicit rather than by inspection of the result.
Prediction
An item is drawn and put back before the next draw.
Predict first
What does that do?
Correct: Makes the two draws independent.
Why: Replacing restores the original situation, so the second draw faces exactly the same counts as the first. Without replacement both counts change and the events become dependent.
Faded example
After one heart has been drawn and not replaced.
Fill in the blanks
P(\text12)=\frac52}___, \text___\frac______}
Why: One heart and one card have both gone, so the numerator and denominator each drop by one. Updating both is what dependence requires.
Explain it
Multiplying works for one case and not the other.
Discussion prompt
Explain to a classmate why independence must be checked explicitly.
Hint: What does the wrong answer look like?
Answer:
Multiplying two probabilities always produces a number, whether or not the events are independent. Nothing about the arithmetic signals a problem.
And the wrong answer is usually close to the right one — one sixteenth against one seventeenth — so it does not look implausible either.
So the check cannot be made by inspecting the result. A good explanation stresses asking whether the first event changes the second's situation, before multiplying rather than after.
Section
Section 4
Concept
An event and its opposite together cover every possibility, so their probabilities sum to one — which means either can be found from the other.
At least one is the classic case: the complement of at least one is none at all, which is a single scenario rather than many. Counting the complement replaces a long case-by-case sum with one computation.
Figure (svg): A card showing the complement rule, with the probability of an event and its opposite summing to one
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1375-1376
Picture it
Two regions filling everything.
Figure (svg): A card showing the complement rule, with the probability of an event and its opposite summing to one
The bottom line names when to reach for it. Questions asking for at least one are almost always easier through the complement.
Worked example
The complement is a single scenario.
\[ \text{Four coins are tossed. What is the probability of at least one head?} \]
Identify the complement
Why: No heads at all.
Count that scenario
Why: One outcome of sixteen.
\[ \frac{1}{16} \]
Subtract from one
Why: The complement rule.
\[ 1 - \frac{1}{16} \]
Simplify
Why: The answer.
\[ \frac{15}{16} \]
Figure (svg): A card showing the complement rule, with the probability of an event and its opposite summing to one
\[ \tfrac{15}{16} \]
Verify: consider the direct route
Why: Directly would mean counting exactly one head, exactly two, exactly three and exactly four, then adding — four combination computations. The complement needed one, which is why the rule is worth reaching for.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1375-1376
Prediction
The event is getting at least one head in several tosses.
Predict first
What is its complement?
Correct: No heads at all.
Why: At least one means one or more, so its opposite is none. That single scenario is far easier to count than summing every case with one or more heads.
Worked example
The complement is far simpler.
\[ \text{Three dice are rolled. What is the probability at least two show the same number?} \]
Identify the complement
Why: All three different.
Count the total outcomes
Why: Six each.
\[ 216 \]
Count the complement
Why: A descending product.
\[ 6 \times 5 \times 4 = 120 \]
Subtract from one
Why: The complement rule.
\[ 1 - \frac{120}{216} \]
Figure (svg): The solution to Worked example a birthday-style question shown as a ladder of expressions, one row per legal move
\[ \tfrac{4}{9} \]
Verify: consider the direct route
Why: Directly would mean counting exactly two matching and all three matching as separate cases, each with its own combination work. The complement is a single descending product, which is a substantial saving.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1376-1376
Error analysis
A student computes an at-least-one probability.
Annotate
On: \( P(\text{at least one head in }4)=P(\text{exactly one head})=\tfrac{4}{16} \)
At least one means one or more, which is nearly everything. Reading it as exactly one gives an answer several times too small, and the complement route makes the correct computation the shorter one.
Faded example
With the complement's probability one sixteenth.
Fill in the blanks
P=1-\frac1615}=\frac___}___
Why: The event and its complement sum to one, so subtracting gives the answer. Counting the single complementary scenario replaced four separate case computations.
Sorting
It usually is for at-least questions.
Sort into buckets
Sort each event.
Explain it to yourself
It is not always easier.
Discussion prompt
Explain what makes it worth using.
Hint: Compare the two counts.
Answer:
It is worth using when the complement is easier to count than the event itself, which is a comparison to make before starting.
At least one is the classic case: the event covers many scenarios and its complement covers exactly one — none at all.
For an exact outcome the reverse holds: the event is one case and the complement is everything else. A good explanation notes that the rule is always valid and the question is only whether it saves work.
Section
Section 5
Concept
Each rule answers a differently worded question, and identifying the wording is what selects the rule before any counting happens.
The words are reliable signals but the structural checks still have to be made: whether the events overlap for a union, and whether they are independent for a product. The word identifies the rule and the check validates it.
| the question says | the rule |
|---|---|
| either, or | the union rule, subtracting the overlap |
| both, and | multiplication, after checking independence |
| at least one | the complement rule |
| not | the complement rule |
| a single event | a ratio of counts |
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1370-1376
Picture it
Words select the rule; this decides how to apply it.
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
Both readings are needed. The word 'and' says to multiply, and the independence check says whether to multiply the stated probabilities or updated ones.
Worked example
A union of dependent products.
\[ \text{Two cards without replacement. Find the probability both are hearts or both are spades.} \]
Find the first probability
Why: Dependent draws.
\[ 13 / 52 \times 12 / 51 \]
Find the second
Why: The same by symmetry.
Check for overlap
Why: Cannot be both suits.
Add
Why: No subtraction needed.
Figure (svg): A contrast between independent events, where probabilities multiply, and dependent ones, where they do not
\[ \tfrac{2}{17} \]
Verify: check both structural decisions
Why: The two draws are dependent, so the second probability was updated; the two events are exclusive, so no overlap was subtracted. Both checks were needed and each affected the answer differently.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1372-1376
Matching
Each phrase signals one approach.
Match the pairs
Why: The wording reliably selects the rule, but each rule then needs its own structural check — overlap for a union and independence for a product. Both readings are required.
Worked example
The wording selects the rule.
\[ \text{What structure does 'at least one of the three is defective' have?} \]
Spot the phrase
Why: At least one.
Identify the complement
Why: None defective.
Note what that needs
Why: Three non-defective draws.
Check independence
Why: With or without replacement.
Figure (svg): The solution to Worked example read the structure first shown as a ladder of expressions, one row per legal move
\[ 1-P(\text{none defective}) \]
Verify: note that both readings were needed
Why: The phrase selected the complement rule and the inner computation still required the independence check. Rules combine in most real questions, so identifying one structure rarely finishes the reading.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1374-1376
Trap
\[ \text{the question says 'and', so multiply the two probabilities} \]
Apply the rule the word indicates
Why: The independence check is skipped.
For dependent events the second probability needed updating first.
The word selects the rule and the check validates it. 'And' means multiply, but which values to multiply depends on independence.
Similarly 'or' means the union rule, but whether the subtraction term is needed depends on whether the events overlap.
Both readings are required, and skipping the second gives an answer that is close enough to look right.
Prediction
A question says both events happen.
Predict first
What must you verify before multiplying?
Correct: Whether the events are independent.
Why: The word selects multiplication but not which values to multiply. Dependent events require the second probability to be updated for what the first changed, and the arithmetic gives no warning either way.
Sorting
Each rule has its own.
Sort into buckets
Sort each situation.
Explain it
Words select rules and checks validate them.
Discussion prompt
Explain the two stages to a classmate.
Hint: What does each stage decide?
Answer:
The wording selects which rule applies: 'or' for the union, 'and' for a product, 'at least one' for the complement.
But each rule then needs a structural check — whether the events overlap, or whether they are independent — which decides how the rule is applied rather than which one it is.
Skipping the second stage gives an answer that is usually close enough to look right. A good explanation stresses that both readings are needed and that the arithmetic never signals which case applied.
Comparison
Fill the blanks from memory. Each answers a differently worded question.
Comparison matrix
| union | multiplication | complement | |
|---|---|---|---|
| signal word | or | and | at least one, or not |
| the operation | add | multiply | subtract from one |
| the check needed | do the events overlap | are they independent | none |
| if the check fails | subtract the overlap | update the second probability | not applicable |
The third row is what separates knowing the rules from applying them. Each has a condition, and the arithmetic gives no warning when one is violated.
Pattern
Five steps, and the middle two are the structural checks.
Steps 3 and 4 are the ones the arithmetic cannot catch. A wrong answer from a skipped check is usually close to the right one, which is what makes them worth doing explicitly.
OpenStax Algebra and Trigonometry 2e, §13.7 Probability §13.7
Check
The union rule.
Check your understanding
Why is the probability of both subtracted when combining two events with or?
Answer: A
Why: Shared outcomes appear in both individual counts, so adding counts them twice. Subtracting the overlap once leaves every outcome counted exactly once, which is what the union requires.
Check
Independence.
Check your understanding
Two cards are drawn without replacement. Are the draws independent?
Answer: A
Why: Removing a card changes both the favourable count and the total for the second draw, so the second probability must be updated before multiplying. Replacing the card would restore independence.
Check
The complement rule.
Check your understanding
What is the complement of getting at least one head in four tosses?
Answer: A
Why: At least one means one or more, so its opposite is none. That single scenario is far easier to count than summing the cases with one, two, three and four heads.
Real world
Medical test interpretation depends on getting dependence right.
Discussion prompt
A test is 99 per cent accurate. Why does a positive result not mean a 99 per cent chance of having the disease?
Hint: What else affects the answer?
Answer:
The test's accuracy and having the disease are not independent — the whole point of the test is that the result depends on the condition.
What matters is how many of the positive results come from people who actually have the disease, and that depends on how rare the disease is.
For a rare disease, most positives can be false ones simply because there are far more healthy people to produce them. The counting has to include the whole population, which is why this is a probability calculation rather than a reading of the accuracy figure — and why the intuitive answer is often badly wrong.
Commit first
State your confidence along with your answer.
Predict first
Why must independence be checked before multiplying probabilities?
Correct: Because dependent events need the second probability updated first.
Why: Multiplying the stated probabilities is correct only when the first event leaves the second's situation unchanged. Otherwise the second must be recomputed for the altered counts, and the arithmetic gives no indication which case applies.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate multiplied two probabilities for cards drawn without replacement. Explain the error.
Hint: What changed after the first draw?
Answer:
Drawing the first card removed it, so the second draw faces fifty-one cards rather than fifty-two — and one fewer of whatever suit was drawn.
So the second probability is not the same as the first and cannot simply be multiplied in unchanged. It has to be updated for what the first draw removed.
The dangerous part is that the wrong answer is close — one sixteenth against one seventeenth. A good explanation stresses that the arithmetic never signals the problem, so the check has to happen before multiplying rather than by inspecting the result.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the costliest error lives, since the wrong answer is close enough to look plausible. The fourth is the technique that most often turns a long computation into a short one.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write probability as a ratio and note the assumption it requires. Beside it, draw two overlapping circles and use them to explain the union rule's subtraction. Underneath, write the multiplication rule with its independence condition and one worked dependent example. Finish with the complement rule and an at-least-one question solved by it.
If your two structural checks are written as questions to ask rather than rules to recall, the section's two costly errors are both accounted for.
Recap
Five things, and two of them are checks the arithmetic cannot make.
| if you remember one thing | it should be this |
|---|---|
| about the ratio | it needs the outcomes to be equally likely |
| about unions | shared outcomes get counted twice unless subtracted |
| about products | check independence; without replacement means dependent |
| about at least one | count the complement, which is a single scenario |
Chapter 12 closes the course by looking forward to calculus: limits, continuity, rates of change and the derivative — the questions the whole of precalculus has been preparing for.
OpenStax, Precalculus, §11.7 Probability §11.7, pp. 1366-1376 — everything on these slides traces back here
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