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How to Solve Probability Questions on the SAT

Probability questions feel like guessing — until you draw the branches. A tree diagram makes SAT probability mechanical: multiply along each branch, add across outcomes, and use the "1 minus" trick for "at least one."

Math · Probability & conditional probability2:10Published July 5, 2026

On YouTube: SAT Probability: Two Rules, Every Question

What this lesson covers

Draw the tree; multiply probabilities along a branch
Without replacement changes the second fraction
All outcomes add to 1
The complement: P(at least one) = 1 − P(none)
Practice: P(at least one red)

Questions worked in the video

0:29A bag has 2 red and 3 blue marbles. Draw 2 without replacement — P(both red)?
1:09Same bag — P(at least one red)?

Worked examples

The questions the video works, written out: the setup, each step, the answer and the trap.

0:29Example 1

A bag has 2 red marbles and 3 blue marbles. You draw two marbles without replacing the first. What is the probability that both are red?

First draw: 2 of the 5 marbles are red, so $\displaystyle P(\text{red first}) = \frac{2}{5}$ .
Second draw, with one red gone: 4 marbles remain and only 1 is red, so the chance of red now is $\displaystyle \frac{1}{4}$ . This is the fraction the no-replacement rule changes.
Along one branch of the tree, multiply: $\displaystyle \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$ .
Check by counting: there are $5 \times 4 = 20$ equally likely ordered pairs of draws, and only $2 \times 1 = 2$ of them are red then red, so $\displaystyle \frac{2}{20} = \frac{1}{10}$ .

Answer: $\displaystyle \frac{1}{10}$

Both red means red AND red, and AND along a branch multiplies, because the second draw only happens after the first has already come up red. The trap answer is $\displaystyle \frac{4}{25}$ from using $\displaystyle \frac{2}{5}$ twice, which forgets that the first red marble left the bag; a second trap is adding the two fractions to get $\displaystyle \frac{13}{20}$ .

0:49Example 2

Same bag, same two draws without replacement. Find the probability of each of the four possible outcomes (red-red, red-blue, blue-red, blue-blue) and show that they add to 1.

Red then red: $\displaystyle \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ .
Red then blue: after a red leaves, 3 of the 4 remaining are blue, so $\displaystyle \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10}$ .
Blue then red: $\displaystyle \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ .
Blue then blue: after a blue leaves, 2 of the 4 remaining are blue, so $\displaystyle \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ .
Add across the outcomes: $\displaystyle \frac{1}{10} + \frac{3}{10} + \frac{3}{10} + \frac{3}{10} = \frac{10}{10} = 1$ . Every possible result is accounted for.

Answer: RR = 1/10, RB = 3/10, BR = 3/10, BB = 3/10, total 1

The four branches are the only things that can happen and no two of them can happen at once, so their probabilities are combined with OR, which means adding, and a complete set of outcomes must total 1. That total is your error check: if the branches do not sum to 1, a second-draw fraction was written with the wrong count or the wrong denominator. Notice RB and BR come out equal; the order differs but the arithmetic is symmetric.

1:09Example 3

Same bag, two draws without replacement. What is the probability that at least one marble is red?

At least one red covers three branches: RR, RB and BR. The only outcome it leaves out is BB, no red at all.
Use the complement: $P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - P(BB)$ .
From the tree, $\displaystyle P(BB) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ , so $\displaystyle 1 - \frac{3}{10} = \frac{7}{10}$ .
Check the long way: $\displaystyle \frac{1}{10} + \frac{3}{10} + \frac{3}{10} = \frac{7}{10}$ . Same answer.

Answer: $\displaystyle \frac{7}{10}$

At least one and none are opposites that together cover everything, so subtracting the single no-red branch from 1 is faster and safer than adding three branches. The trap answer is $\displaystyle \frac{3}{10}$ , stopping at P(no red) and forgetting the 1 minus; another is $\displaystyle \frac{4}{10}$ from adding only RR and RB and dropping BR.

Chapters

Lesson transcript

The narration of the video, word for word, under its chapter headings.

0:00Stop guessing — draw the branches

Welcome to SAT Climb. Probability questions feel like guessing — until you draw the branches. A tree diagram turns 'I think' into 'I know.'

0:16The setup: 2 red, 3 blue

A bag has 2 red marbles and 3 blue. You draw 2, without replacing the first. What's the probability both are red?

0:29Multiply along the branch

Draw the tree. First pick: red is 2 of 5. Then, with one red gone, red is 1 of 4. Multiply along the branch: 2/5 × 1/4 = 1/10. That's both red.

0:49Add across + the 1-minus trick

Every branch multiplies; every outcome adds. All four add to 1. Shortcut: for 'at least one red,' take 1 minus the no-red case. 1 − 3/10 = 7/10.

1:09Your turn

Your turn. Same bag. What's the probability of at least one red? Use the shortcut: 1 minus the probability of no red.

1:35Three traps

Three traps. Adding along a branch instead of multiplying. Forgetting it's without replacement, so the 2nd fraction changes. And doing 'at least one' the long way instead of 1 minus.

1:56Recap

Probability: solved. Draw the branches, and the numbers follow. Start free at satclimb.com.

Read the written version: the Probability & conditional probability strategy guide, then try 25 hard Probability & conditional probability questions with full explanations. Both are free, no account needed.

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