Mistake Master
Whose world, which operation
A probability is a fraction with a story: favorable outcomes over the world the question sets. The arithmetic is small. The points slip away when a condition fails to shrink the denominator, when AND and OR trade operations, and when the complement or a raw count stands in for the probability asked. Fix the world, pick the operation, name the event.
§1
What this topic is about
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A probability is favorable over total, for one named event. The SAT builds its traps around three moves: the denominator a condition demands, the operation that combines two events, and the difference between a probability and its look-alikes: the complement, a raw count, the odds.
§2
The condition names the denominator
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"If a senior is chosen" shrinks the world to the seniors: they become the denominator, and only seniors can appear in the numerator. Ignoring the condition, or conditioning on the wrong group, are the two classic failures.
- The group named after "if" (or "given") is the new total.
- No condition stated: the whole table is the denominator.
- P(science given senior) and P(senior given science) are different questions with different denominators.
Worked example. Of $100$ seniors, $70$ chose science (survey of $200$). If a senior is chosen at random, what is P(science)?
The condition makes the denominator $100$, the seniors.
$$\dfrac{70}{100} = \dfrac{7}{10}.$$ Dividing by $200$ ignores the condition; dividing by the $130$ science choosers answers the reversed question.
§3
AND multiplies, OR adds, at-least-one flips
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Two independent events both happening: multiply. Two events that cannot both happen, either one: add. At least one out of several: go through the complement, the chance that none happen.
- Independent AND: $P(A) \cdot P(B)$. Adding here can even exceed $1$.
- Exclusive OR: $P(A) + P(B)$.
- At least one: $1 - P(\text{none})$.
Worked example. Rain has probability $0.3$ each of two independent days. P(at least one rainy day)?
No rain at all: $0.7 \cdot 0.7 = 0.49$.
$$1 - 0.49 = 0.51.$$ Adding $0.3 + 0.3 = 0.6$ double-counts the both-rainy case; multiplying $0.3 \cdot 0.3$ answers BOTH days, not at least one.
§4
Without replacement, the pool shrinks
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When a draw is not returned, the second draw faces a smaller pool: both the favorable count and the total drop by what the first draw took.
Worked example. A box holds $4$ red and $6$ blue pens; two are drawn without replacement. P(both red)?
$\dfrac{4}{10}$.
$$\dfrac{4}{10} \cdot \dfrac{3}{9} = \dfrac{2}{15}.$$ Multiplying $\dfrac{4}{10}$ twice quietly puts the first pen back.
§5
A probability, not its look-alikes
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Every probability question spawns a family: the complement, the counts, the odds, the expected number. The choices will contain them all.
- A probability's denominator is the WHOLE space; odds compare favorable to unfavorable.
- P(not A) is $1 - P(A)$; check which event the question names.
- Expected count $=$ probability $\times$ number of trials.
Let the condition set the denominator, match the operation to AND, OR, or at-least-one, shrink the pool when draws are not replaced, and report the probability of the exact event named.
§6
Three patterns that cost real points
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Three patterns recur on probability questions. They are the same ones the diagnostic routes on.
The denominator ignores the condition.
P(science given senior) gets computed over all $200$ students, or the condition flips to the outcome group and answers the reversed question.
Fix. The if-clause names the denominator. Rewrite the question as "out of [that group], how many..." before dividing.
The wrong operation combines the events.
Both-and gets added, either-or gets multiplied, at-least-one skips the complement, or a without-replacement draw keeps the original counts.
Fix. Say the structure out loud: both (multiply), either (add, if exclusive), at least one (complement). And every unreturned draw shrinks the next denominator.
A look-alike answers instead.
The complement, a raw count, or the odds gets reported where the probability was asked, or the probability where an expected COUNT was.
Fix. Check the event (not its opposite), check the denominator (the whole space), and check the type: probabilities live in $[0, 1]$; expected counts do not.
Ten quick checks across the patterns: conditions and denominators, AND versus OR versus at-least-one, draws without replacement, complements, and expected counts. Pick or type your answer, then check. Progress is saved.