# GRE Probability Without Replacement: Examples | topin

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## Write the pool after every removal

[ETS](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html) includes elementary and conditional probability in GRE Quant. The [Math Review](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-review.pdf) distinguishes independent events from draws where the first outcome affects the next. If an object stays out after selection, the remaining pool has one fewer object. The favourable count may also shrink, depending on what was selected. Writing only the new denominator is therefore an incomplete update.

For a bag with five red and four blue tokens, the probability of red on the first draw is 5/9\. After red is removed, the next red probability is 4/8, while the next blue probability is 4/8\. After blue is removed instead, the next red probability is 5/8\. Conditional probabilities describe the branch you are actually following.

The replacement condition is a favourite lever in [Quantitative Comparison](/gre/quant/quantitative-comparison). A box holds six green and four yellow counters and two are drawn. Quantity A: the probability both are green with replacement. Quantity B: the same probability without replacement. A is (6/10)(6/10) = 9/25 = 0.36; B is (6/10)(5/9) = 1/3\. A is greater, and you can see why without arithmetic: removing a green counter makes a second green less likely. Label the counts in scratch work and state whether replacement is allowed before you multiply anything. The [GRE Quant section guide](/gre/quant) sets out all five question formats.

## Choose multiplication, addition or a complement

Multiply along a sequence because every step on that branch must happen. Add probabilities of distinct sequences when either sequence satisfies the request and they cannot both describe the same outcome. Subtract from one when the excluded event is simpler. These operations are useful only after the event is defined precisely; “two red” and “at least one red” are different events.

The table is our method-selection aid, not a rule about how frequently a form appears on the GRE. In every case the denominator reflects the actual experiment. Ordered draws are convenient for sequential conditions. Unordered selections are convenient when only the final group matters. Do not put unordered favourable groups over ordered total sequences, because those counts are measured in different units.

__Match the event before calculating__
| Requested event                         | Useful model                           | Important condition              |
| --------------------------------------- | -------------------------------------- | -------------------------------- |
| Red then blue                           | Multiply one ordered branch            | Update after the red draw        |
| One red and one blue in either order    | Add red–blue and blue–red branches     | Branches must be disjoint        |
| At least one red                        | 1 − probability of no red              | Count all excluded draws         |
| Exactly two red in a three-object group | Combinations or all matching orders    | Use the same ordering convention |
| Second draw given the first result      | Conditional remaining-pool probability | Use the known first outcome      |

## Original problem: calculate a specified order

Original question: a box contains six green and four yellow counters. Two counters are drawn uniformly at random, without replacement. What is the probability that the first is green and the second is yellow? The first probability is 6/10\. Given that green was removed, four yellow counters remain among nine counters, so the second probability is 4/9.

The required probability is (6/10)(4/9) = 24/90 = 4/15\. Using 4/10 for the second draw would treat the first counter as though it had been returned. Using 3/9 would remove a yellow counter even though the branch says the first draw was green. Both errors can produce plausible fractions, so the branch labels are a better check than size alone.

If the request instead says green then green, the calculation is (6/10)(5/9) = 1/3\. Both numerator and denominator shrink on the second draw. If replacement is introduced and the box is remixed, green then yellow becomes (6/10)(4/10) = 6/25\. The original total is appropriate only in that altered experiment.

## Original problem: include both possible orders

Original question: using the same six-green, four-yellow box, what is the probability that the two drawn counters have different colours? They can be green then yellow or yellow then green. The first branch is 4/15\. The second is (4/10)(6/9) = 4/15\. The branches cannot both occur in the same ordered draw, so add them to obtain 8/15.

Doubling works here because the two branches have equal probability. Do not treat “either order” as a universal instruction to double: with three positions, conditions or different category counts, the number of branches can change. First list the valid orders, then decide whether symmetry allows grouping them. For a small problem, that list is often shorter than correcting an overgeneralised shortcut.

An unordered check chooses one green and one yellow in 6 × 4 = 24 ways. All two-counter groups number 10 choose 2 = 45\. The ratio is 24/45 = 8/15, agreeing with the ordered method. [Combinations and permutations](/articles/gre-combinations-permutations) explains why dividing by order removes repeated representations of the same final group.

## Original problem: use the opposite event

Original question: a box contains three marked cards and seven unmarked cards. Three cards are selected uniformly without replacement. What is the probability that at least one card is marked? Directly calculating one, two and three marked cards requires several cases. The opposite event is no marked cards, meaning all three are selected from the seven unmarked cards.

The no-marked probability is (7/10)(6/9)(5/8) = 7/24\. Therefore at least one marked has probability 1 − 7/24 = 17/24\. A complement is exact because the original event and its opposite exhaust all possibilities without overlap. Do not subtract the probability of exactly one unmarked card: that is not the opposite of at least one marked.

Check the boundary cases before using the formula. If every card is unmarked, the complement produces zero for the original event. If only two unmarked cards exist and three are drawn, no-marked is impossible and the original event has probability one. This test is especially useful when a calculation contains a zero numerator that looks suspicious but actually expresses an impossible branch.

## Original problem: exactly two successes in three draws

Original question: there are five red and three blue beads, and three are selected uniformly without replacement. What is the probability of exactly two red beads? Count final groups: choose two of the five red beads in ten ways and one of three blue beads in three ways. The favourable count is 30\. The total count is 8 choose 3 = 56.

The probability is 30/56 = 15/28\. A sequential check has three orders: red–red–blue, red–blue–red and blue–red–red. Each has probability 5 × 4 × 3 divided by 8 × 7 × 6, equal to 5/28\. Adding all three gives 15/28\. “Exactly” excludes three red, so do not add the all-red branch.

With a condition such as “given that the first bead was blue”, the sample space changes again. The known draw leaves five red and two blue beads. Exactly two red among all three now requires both remaining draws to be red, with probability (5/7)(4/6) = 10/21\. A conditional question should start from the stated information, rather than recalculating the probability that the known condition occurred.

For a final original check, suppose there are only two red beads and three are drawn. The probability of three red is zero, because the third red numerator becomes zero. Do not discard that branch as an arithmetic problem or use a negative favourable count. A zero probability can express a genuine impossibility in the stated pool.

## Check the sample space before checking the arithmetic

Our review routine asks four questions: was selection uniform, was replacement allowed, did order matter, and did the event say exactly or at least? Then verify every remaining-pool numerator and denominator. A probability must lie between zero and one, but that range check is weak by itself. Many incorrectly modelled bag problems still produce a number inside that interval.

Practise changing only one word in the same scenario: “with” to “without” replacement, “then” to “either order”, and “exactly one” to “at least one”. Explain the new event before solving. Later use [topin’s free GRE mock](/gre/practice-test), marked on the official scale, to test whether you retain that discipline among unrelated Quant problems. Our practice sequence is a recommendation.

Keep fractions during cancellation, using the calculator for tedious final arithmetic if necessary. ETS advises against introducing decimals unnecessarily when an answer is requested as a fraction. Follow the [Numeric Entry instructions](/gre/quant/numeric-entry) for the final format. If a result needs rounding, round the completed probability as instructed, rather than rounding each branch before adding.

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## FAQs

Why does the denominator decrease without replacement?

The selected object is removed, leaving one fewer available object for the next draw. The favourable numerator also changes when the removed object belongs to the favourable category for that branch.

Do I always multiply probabilities for two draws?

Multiply conditional probabilities along a specified sequence. If several disjoint sequences meet the event, add their branch probabilities afterwards. Using the original second-draw probability without checking dependence can give the wrong result.

When should I use a complement?

Use it when the opposite event is easier, especially for “at least one”. Subtract the probability of none from one. Make sure the opposite event excludes every outcome in the original event.

Can combinations solve a without-replacement problem?

Yes, when the final group matters and equal-size groups are equally likely. Count favourable groups and total groups using the same unordered convention. An ordered condition is often clearer with sequential probabilities.

Does a known first result need its probability multiplied in?

Not when the question asks for a probability conditional on that known result. Start with the pool remaining after it. Multiply its probability only when the original event includes the first result as an uncertain part of the outcome.

## Sources (checked 5 October 2026)

* [ETS: Quantitative Reasoning overview and calculator guidance (checked 5 October 2026)](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html)
* [ETS: GRE Math Review (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-review.pdf)
* [ETS: GRE Mathematical Conventions (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-conventions.pdf)
* [ETS: Guidelines specific to the on-screen calculator (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/on-screen-calculator-guidelines.pdf)

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