10.4 Complementary Probability & Multi-Stage Events

Complements Are the Short List

Many GMAT probability stems ask for a messy event: at least one, at least two, not all, not both. The complementary event is often a single simple case — none, all, or exactly a full miss. Official Guide Math Review 3.4 expects this rewrite. The live section is still 21 Problem Solving questions in 45 minutes with no calculator, so a one-term complement beats a five-term direct expansion.

P(E) = 1 − P(E^c)

E^c ("not E," the complement) is every outcome where E fails. Because E and E^c cannot overlap and together they fill the sample space, their probabilities add to 1.

Worked example. Fair coin flipped 3 times. P(at least one head).

Direct: 1 head, or 2, or 3: C(3,1)/8 + C(3,2)/8 + C(3,3)/8 = 3/8 + 3/8 + 1/8 = 7/8.

Complement: P(no heads) = P(TTT) = 1/8. Then 1 − 1/8 = 7/8.

The complement is one outcome; the direct count is seven. Prefer the complement whenever "at least one" appears.

Worked example. Two fair dice. P(at least one 6).

P(no 6) = (5/6) × (5/6) = 25/36 (independent rolls). P(at least one 6) = 1 − 25/36 = 11/36.

Direct check: exactly one 6 uses two ordered paths, (1/6)×(5/6) + (5/6)×(1/6) = 10/36. Two 6s: 1/36. Total 11/36. Direct work needed a careful "exactly one" term so the double-6 was not counted twice. The complement never had that problem.

Worked example. Four good parts and two defective in a box. Two parts drawn without replacement. P(at least one defective).

P(none defective) = P(both good) = (4/6) × (3/5) = 12/30 = 2/5. P(at least one defective) = 1 − 2/5 = 3/5.

Combination check: C(4,2)/C(6,2) = 6/15 = 2/5 for both good, so 1 − 2/5 = 3/5. Direct: exactly one defective C(2,1)×C(4,1)/15 = 8/15, plus both defective C(2,2)/15 = 1/15, total 9/15 = 3/5.

"At Least One" Is Not "Exactly One"

Exactly one is a slice. At least one is that slice plus everything above it. If a stem wanted exactly one head in three flips, the answer is 3/8, not 7/8. Read the quantifier.

At least two is a complement of a two-piece event: 1 − P(0) − P(1). You may still complement; you just subtract both small cases.

Worked example. Three independent yes/no items, P(correct) = 1/2 each. P(at least two correct) = 1 − P(0 correct) − P(1 correct) = 1 − 1/8 − 3/8 = 4/8 = 1/2.

That 1/2 also equals P(exactly 2) + P(exactly 3) = 3/8 + 1/8. Symmetric coins make at-least-two equal to at-most-one, which is a coincidence of p = 1/2, not a rule. If P(correct) were 1/3, those would not match: P(0) = (2/3)³ = 8/27 and P(1) = 3 × (1/3)(2/3)² = 12/27, so P(at least two) = 1 − 20/27 = 7/27, while P(at most one) = 20/27.

"Not All" Versus "None"

These are different complements of different events.

• None = every trial fails. For n independent fair-coin heads, (1/2)^n.
• All = every trial succeeds. Also (1/2)^n for a fair coin.
• Not all = 1 − P(all). This includes "none" and every mixed outcome.
• Not none = at least one = 1 − P(none).

Worked example. A 4-question true/false quiz, each guessed independently with two options. P(every answer correct) = (1/2)^4 = 1/16. P(not every answer correct) = 1 − 1/16 = 15/16. P(no answer correct) = 1/16.

15/16 is not 1/16. Testers who swap "not all" for "none" pick the complement of the wrong event.

Worked example with unequal probabilities. Three independent alarms, each fails with probability 1/10 and sounds with probability 9/10.

P(all three fail) = (1/10)³ = 1/1,000. P(at least one sounds) = 1 − 1/1,000 = 999/1,000. P(not all sound) = 1 − (9/10)³ = 1 − 729/1,000 = 271/1,000.

Those are not interchangeable. "At least one sounds" is a safety question. "Not all sound" is a different failure mode.

Sequential Trees

A tree is the slot method with probabilities on the branches. Each full path is an AND (multiply along the path). Disjoint paths that all satisfy the event are an OR (add those path products).

Rules:

1. At every node, the outgoing branch probabilities add to 1.
2. Without replacement, later-branch fractions change.
3. A two-stage or three-stage forward tree is ordinary GMAT probability. Label each branch P(this stage | previous stages) and multiply. You do not need a reverse-inference course to walk a tree in the direction the experiment actually runs.

Worked example. An urn holds 3 red and 2 green. Draw one, do not replace, draw a second. P(second is red).

• Path RR: (3/5) × (2/4) = 6/20
• Path GR: (2/5) × (3/4) = 6/20
• Total: 12/20 = 3/5

Check: P(second is red) = 3/5, same as P(first is red). Without extra information, each draw is equally likely to be any of the five original marbles. The tree still teaches the method, because you had to add two disjoint paths.

Worked example, three stages. A candidate answers three independent questions, P(correct) = 2/3 each. P(at least one correct) = 1 − (1/3)³ = 1 − 1/27 = 26/27.

Worked example, mixed replacement. Two draws from 5 red and 3 blue.

P(same color) = P(RR) + P(BB).

Without replacement:

• RR: (5/8)(4/7) = 20/56
• BB: (3/8)(2/7) = 6/56
• Same color: 26/56 = 13/28

P(different colors) = 1 − 13/28 = 15/28, matching the one-red-one-blue calculation in 10.3.

With replacement:

• RR: (5/8)² = 25/64
• BB: (3/8)² = 9/64
• Same: 34/64 = 17/32
• Different: 1 − 17/32 = 15/32

Replacement changes the answer. Read whether the object goes back.

Geometric probability is out of scope

Stay with counting sample spaces, independent products, dependent sequential products, unions, and complements. Do not build length-or-area probability diagrams. A random point on a segment is not a Focus Edition pattern you should study toward.

A Full Pass on One Multi-Stage Item

A drawer has 4 black socks and 2 white socks. Two socks are drawn without replacement.

• P(both black) = (4/6)(3/5) = 12/30 = 2/5
• P(both white) = (2/6)(1/5) = 2/30 = 1/15
• P(a matching pair) = 2/5 + 1/15 = 6/15 + 1/15 = 7/15
• P(not a matching pair) = 1 − 7/15 = 8/15 (this is one black and one white)
• P(at least one black) = 1 − P(both white) = 1 − 1/15 = 14/15
• P(at least one white) = 1 − P(both black) = 1 − 2/5 = 3/5

Every line is either a path product, a sum of path products, or a complement. That is the whole GMAT probability story: count or multiply to a simple event, then take 1 minus that event when the stem says "at least," "not all," or "not none."

Traps on Quant

• Computing P(exactly one) when the stem said at least one.
• Computing P(none) when the stem said not all.
• Forgetting to subtract P(1) when the complement of "at least two" is "0 or 1."
• Using (1/2)^n for "at least one" instead of 1 − (1/2)^n.
• Changing a without-replacement tree into a constant-factor independent product.
• Treating a "given the second marble is red" prompt as a forward P(second is red). If the stem conditions on a later stage, keep only the paths that match that stage, then take the share of those paths you want.