SAT · Problem-Solving and Data Analysis · TOPIC LESSON
Count the right outcomes, condition on the right group, and recover missing information.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
The sample space is the collection of possible outcomes. For equally likely outcomes, P(event) = favorable count/total count. A probability lies between 0 and 1 inclusive. Relative frequency is the observed proportion in a dataset; it can estimate a probability but is not a guarantee about every future sample.
Do not assume outcomes are equally likely simply because they have different names. A spinner sector twice as large has twice the probability only under a uniform-angle spin model.
02 / UNDERSTAND & APPLY
The complement of event A consists of all outcomes not in A, so P(not A) = 1 − P(A). This is useful when directly counting A is harder. Categories must cover the whole reference group without overlap before their counts can be added as a simple total.
ORIGINAL WORKED EXAMPLE
A box contains 7 amber tokens, 5 blue tokens and 8 gray tokens, otherwise identical. One token is selected uniformly. What is P(not blue)?
Answer 3/4
03 / UNDERSTAND & APPLY
An interior cell counts individuals with both its row and column categories. Row and column totals are marginal counts. The grand total is the reference group for an unconditional probability. Avoid adding row totals and column totals together, which counts each person twice.
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ORIGINAL WORKED EXAMPLE
In the displayed 60-person table, find the probability that a uniformly selected person is both a morning attendee and a cyclist.
Answer 1/5
04 / UNDERSTAND & APPLY
Conditional probability P(A given B) restricts attention to outcomes in B. Its denominator is the count in B, and its numerator is the count in both A and B. In probability notation, P(A given B) = P(A and B)/P(B), provided P(B) > 0.
P(A given B) and P(B given A) usually differ because their reference groups differ. Read the words after “given” first to identify the denominator.
ORIGINAL WORKED EXAMPLE
Using the table, compare P(cyclist given morning) and P(morning given cyclist).
Answer 2/5 and 3/8, respectively
05 / UNDERSTAND & APPLY
For inclusive “A or B,” count outcomes in A, plus outcomes in B, minus those in both because they were counted twice. Thus P(A or B) = P(A) + P(B) − P(A and B). Mutually exclusive events have no overlap. Independent events are a different idea: knowing one does not change the probability of the other.
In the displayed table, morning or cyclist includes 30 + 32 − 12 = 50 of 60 people, giving 5/6. Simply adding 30 and 32 would double-count the morning cyclists.
06 / UNDERSTAND & APPLY
If a point is selected uniformly by area within a region, probability of landing in a subregion is subregion area/total area. Similar reasoning applies to uniform selection along a length. The region must be contained in the sample region and boundary conventions have no effect on continuous area probability because boundary lines have zero area.
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ORIGINAL WORKED EXAMPLE
A point is chosen uniformly from a 10-by-6 rectangle. A shaded 4-by-3 rectangle lies inside it. Find the shaded probability.
Answer 1/5
07 / UNDERSTAND & APPLY
Write favorable/total = given probability. If the unknown count contributes to the total, include it there too. The result must be a nonnegative integer when it counts objects. A displayed rounded percentage may not uniquely determine an exact count unless the question gives enough precision.
ORIGINAL WORKED EXAMPLE
A bag contains 9 striped tokens and n plain tokens. A uniformly selected token is plain with probability 2/5. Find n.
Answer 6 plain tokens
08 / UNDERSTAND & APPLY
For independent events, P(A and B) = P(A)P(B). In general the multiplication rule is P(A and B) = P(A)P(B given A). Drawing without replacement changes the second sample space, so independence usually fails.
A calculator can simplify fractions and divide counts, but cannot choose the denominator for you. Check whether the question selects from the whole group, a restricted group, or a sequence of changing groups.
09 / UNDERSTAND & APPLY
Assume uniform selection only when the question says so.
ORIGINAL PRACTICE · 1
3/10
There are 40 students total, so 12/40 = 3/10.
ORIGINAL PRACTICE · 2
0.73
Complement probabilities sum to 1.
ORIGINAL PRACTICE · 3
3/8
The given group is the 24 swimmers, giving 9/24.
ORIGINAL PRACTICE · 4
16/25
Count the union: 20 + 18 − 6 = 32; divide by 50.
ORIGINAL PRACTICE · 5
24
0.3 × 80 = 24.
ORIGINAL PRACTICE · 6
8
4/(4 + n) = 1/3 gives 12 = 4 + n.
ORIGINAL PRACTICE · 7
1/5
Uniform-area probability is 9/45.
ORIGINAL PRACTICE · 8
No
Their denominators are different groups, so reversing the condition usually changes the result.
ORIGINAL PRACTICE · 9
1/16
Independence permits multiplication: (1/4)(1/4).
KEEP THE BIG IDEAS
READ WITH CONTEXT
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