GRE Probability: Counting Outcomes and Solving Practice Problems

GRE probability questions are usually built from a modest set of principles: define possible outcomes, decide which are favorable, and count carefully. The arithmetic may be simple while the sample space is easy to misread. A good solution begins by stating what one outcome means and whether outcomes are equally likely.

The basic probability fraction

When outcomes are equally likely, probability equals favorable outcomes divided by total possible outcomes. A fair six-sided die has six equally likely faces. The chance of rolling an even number is 3/6 = 1/2 because the favorable faces are 2, 4, and 6. Probability ranges from 0 to 1; multiplying by 100 expresses it as a percentage.

The equally likely condition matters. If a spinner has regions of different sizes, the regions are not necessarily equally likely. Counting colored regions would be insufficient unless each region has equal area or the problem otherwise states equal likelihood. The denominator must represent outcomes with equal probability, not simply the number of labels in a picture.

Define outcomes at the level needed for the question. If you draw two cards and care about their order, drawing red then black differs from black then red. If the question only asks how many cards of each color were drawn, those sequences may describe the same unordered result. A clear definition avoids counting the same event inconsistently.

Complementary events

An event and its complement cover all possibilities and cannot happen together. Therefore P(not A) = 1 − P(A). This is especially useful for ‘at least one’ questions. Instead of counting every way to get one or more successes, calculate one minus the probability of getting none.

If a fair coin is flipped four times, the probability of at least one head is 1 − P(no heads) = 1 − (1/2)⁴ = 15/16. Directly counting outcomes with one, two, three, or four heads is possible, but the complement is shorter and less prone to omission.

‘And’ and ‘or’ relationships

For independent events A and B, P(A and B) = P(A) × P(B). Independence means learning that one event occurred does not change the probability of the other. Separate coin flips are independent; drawing two objects without replacement generally is not, because the first draw changes what remains.

For mutually exclusive events, which cannot occur together, P(A or B) = P(A) + P(B). If events can overlap, subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). Adding probabilities without subtracting a shared outcome counts that outcome twice.

The words ‘and’ and ‘or’ are not enough by themselves to select a formula. Ask whether the events are independent, mutually exclusive, or overlapping. A number can be even and greater than 3, so those events overlap. On a die, the outcomes 4 and 6 satisfy both.

Conditional probability and restricted samples

Conditional probability asks for the chance of A given that B is known. The sample space shrinks to outcomes satisfying B. In a standard deck, if a card is known to be a face card, the chance it is a king is 4/12 = 1/3, because the relevant sample space contains 12 jacks, queens, and kings. The chance of drawing a king from the full deck is 4/52; that is a different question.

For sequential draws without replacement, update the denominator and favorable count after the first draw. If a bag contains 3 red and 2 blue counters, the probability of two red draws without replacement is (3/5)(2/4) = 3/10. The second red probability is 2/4 because one red counter has already been removed.

With replacement, the contents return to their original state before the next draw. The same two-red probability is (3/5)(3/5) = 9/25. These procedures are different experiments. Mark ‘with replacement’ or ‘without replacement’ before multiplying.

Counting outcomes without double counting

If a process has successive independent choices, multiply the number of choices at each stage. A code with two letters followed by one digit has 26 × 26 × 10 possible codes if repetition is allowed. If letters cannot repeat, the count becomes 26 × 25 × 10. The phrase ‘repetition allowed’ changes the second choice count.

Order matters for arrangements. Placing three different books on a shelf can be done in 3 × 2 × 1 = 6 orders. If the question asks only which three books are selected from a larger group, order does not matter. In that setting, counting each arrangement as a different selection overcounts.

For small groups, list the possibilities systematically. For larger counts, use combinations: choosing k objects from n distinct objects is n!/[k!(n − k)!]. Permutations count ordered arrangements: n!/(n − k)!. You do not need a formula when a short organized list or multiplication rule is clearer.

A common error is to count favorable cases with order but count total cases without order. Keep the sample-space convention consistent. If order matters in the numerator, it must matter in the denominator too; if order does not matter, count both the favorable and total selections without order.

Original GRE-style practice

Question 1: complement

A fair six-sided die is rolled three times. What is the probability of getting at least one 6?

1. 1/6
2. 1/3
3. 91/216
4. 125/216

Answer: C, 91/216. The chance of no 6 on one roll is 5/6. For three independent rolls, it is (5/6)³ = 125/216. The complement is 1 − 125/216 = 91/216. The tempting 1/2 is not justified; three chances do not make the probability exactly one half.

Question 2: without replacement

A box contains 4 green and 3 yellow tokens. Two tokens are selected one after another without replacement. What is the probability both are green?

1. 1/7
2. 2/7
3. 3/7
4. 4/7

Answer: B, 2/7. The first green probability is 4/7. After drawing one green, 3 green tokens remain among 6 total, so the second probability is 3/6 = 1/2. Multiply: (4/7)(1/2) = 4/14 = 2/7. The favorable ordered draws are 4 × 3, out of 7 × 6 total ordered draws, also 12/42 = 2/7.

Question 3: overlap in an ‘or’ event

One integer is selected at random from 1 through 12. What is the probability it is a multiple of 2 or a multiple of 3?

1. 1/2
2. 2/3
3. 3/4
4. 5/6

Answer: B, 2/3. There are 6 multiples of 2 and 4 multiples of 3, but 2 numbers (6 and 12) are in both groups. Favorable values are 6 + 4 − 2 = 8, so probability is 8/12 = 2/3. Adding the two group sizes without subtracting their overlap would count 6 and 12 twice.

Question 4: order and repetition

A three-character code uses digits 0 through 9, and repetition is allowed. What is the probability that all three digits are the same?

1. 1/100
2. 1/10
3. 3/10
4. 1/3

Answer: A, 1/100. There are 10 choices for each of three positions, or 1,000 equally likely codes. Ten codes have all digits the same: 000, 111, through 999. The probability is 10/1,000 = 1/100. Once the first digit is chosen, the remaining two must match it, so there is one favorable continuation out of 100 possible pairs.

Question 5: conditional probability

A club has 12 members, 7 of whom are seniors. Among the seniors, 4 study biology. If a randomly selected member is known to be a senior, what is the probability that the member studies biology?

1. 1/3
2. 4/7
3. 4/12
4. 7/12

Answer: B, 4/7. The condition restricts the sample space to the 7 seniors. Four satisfy the biology condition, so the probability is 4/7. The fraction 4/12 is the probability of both being a senior and studying biology from the full club, not the requested conditional probability.

Expected counts are not probabilities

A probability can be used to estimate a long-run count. If an event has probability 0.2 on each of 50 independent trials, the expected number of occurrences is 50 × 0.2 = 10. That does not guarantee exactly 10 occurrences in 50 trials. Expected value is an average over repeated comparable experiments.

Some GRE problems ask how many outcomes satisfy a condition rather than for probability. Count the favorable outcomes directly. If asked for a probability, divide by the total equally likely outcomes. Avoid introducing probability when the question only asks for a count.

Probability and quantitative comparison

In quantitative comparison, simplify fractions or compare complementary events before converting to decimals. For instance, compare 7/12 and 3/5 by cross multiplication: 7 × 5 = 35, while 3 × 12 = 36, so 7/12 is smaller. This can be faster and exact.

Check boundary cases. An impossible event has probability 0 and a certain event has probability 1. If your answer is negative or greater than 1, the setup is wrong. If two disjoint events exhaust all outcomes, their probabilities should add to 1.

A disciplined solution checklist

1. Define one outcome and write the complete sample space or a counting rule.
2. Identify whether outcomes are equally likely and whether order matters.
3. Mark whether events overlap, are mutually exclusive, or are independent.
4. For a condition, restrict the sample space before calculating.
5. For ‘at least one,’ look for the complement of ‘none.’
6. Check that the final probability lies between 0 and 1 and matches the requested event.

How to practice probability

Start with small samples you can list: a few coin flips, a short number set, or a bag with a handful of tokens. Write all outcomes in an organized table or tree. Then compare that list with a multiplication or combination calculation. This exposes whether order matters and whether you have counted every case once.

Once the setup is reliable, mix problem types without labels. A practice set titled ‘independent events’ has already told you which rule to use. Real questions require you to infer it. After each miss, record the sample space, replacement condition, overlap, and denominator you chose. Those notes are more useful than memorizing a list of formulas.

ETS's free math review supports foundational quantitative study. The current GRE has 27 Quantitative Reasoning questions across two sections, but ETS does not guarantee a fixed number of probability items. Prepare the reasoning broadly and use original examples to test whether you can recognize the relevant structure in unfamiliar settings.