SAT probability: tables, conditions, and sample spaces

01Name the eligible group first

A bag contains 5 red tiles, 3 blue tiles, and 2 green tiles. If one tile is selected at random, the probability it is blue is $3/10$. There are 3 favorable tiles and 10 eligible tiles.

The wording "selected at random" matters because the count method assumes each eligible item has the same chance of selection. If the question describes a different selection process, use the probabilities it gives instead.

Before you calculate, finish this sentence: "The selection is being made from ___." That blank names the denominator. When a table or condition appears, the denominator may be a single row rather than the entire sample.

02Read a two-way table in both directions

Suppose a survey classifies students by grade and whether they walk to school. The table contains counts, so row and column totals can be added directly.

03Joint and conditional probabilities use different denominators

If one surveyed student is selected at random, the probability the student is a senior who walks is $12/100=0.12$. Both descriptions must be true, and the student was selected from all 100 students.

If a senior is selected at random, the probability that student walks is $12/40=0.30$. The condition "a senior" restricts the eligible group to the 40 seniors. The probability of being a senior given that the student walks is $12/30=0.40$.

Those last two probabilities reverse the condition. They use the same intersection count of 12 but different denominators. Read the sentence all the way through before looking for a cell. A symbol such as $P(W\mid S)$ means probability of W given S; the group after the vertical bar supplies the denominator.

04Handle "or" without counting the overlap twice

In the same table, the probability a randomly selected student is a senior or walks is $\frac{40+30-12}{100}=0.58$. Adding all seniors and all walkers counts the 12 senior walkers twice, so subtract the overlap once.

In general, add both probabilities and subtract the overlap: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here the union sign means either event and the intersection sign means both. If the events cannot both happen, the overlap is zero. A single tile cannot be both red and blue in the bag example, so the probability is $(5+3)/10=0.8$.

"Or" includes the overlap unless the problem explicitly excludes it. The table makes this easier to see: include every eligible cell meeting at least one condition, once.

05Use the complement when it saves counting

The probability a tile is not green is $1-P(\text{green})=1-2/10=0.8$. You could also add red and blue tiles. Choose the shorter count, then check that the result lies between zero and one.

For two draws without replacement from the 10-tile bag, the probability both tiles are red is $\frac{5}{10}\cdot \frac{4}{9}=\frac{2}{9}$. After the first red tile is removed, 4 red tiles remain out of 9 total. With replacement, the second fraction would be $5/10$ again, giving $1/4$.

Multiply the first probability by the second probability under the stated condition. Don't assume the draws are independent when removing an item changes the bag. Equally, don't subtract an item when the question says it is replaced.

06Separate a probability from a prediction

If a random sample finds that 30 of 100 students walk, 30% is the observed sample proportion. Applying that rate to a population of 800 students gives an estimate of $0.30(800)=240$ walkers, not proof that exactly 240 walk.

When a question asks for a probability within the sample, use the table's counts. When it asks for an estimated population count, use the proportion and the population size. A representative sampling method matters for that estimate; a survey of only the walking club would not justify applying its rate to the school.

Practice probability and conditional probability in the Math bank. For each missed problem, write the denominator you used and the denominator the wording required. That comparison gives you a specific fix.