10.3 Probability Rules: Independent & Dependent Events

Equally Likely Outcomes

Official Guide Math Review 3.4 puts probability on the same map as counting. On Focus Edition Quantitative Reasoning, probability is still Problem Solving: 21 questions, 45 minutes, no calculator. GMAC tests fractions you can reduce by hand, not a statistics-course toolkit. You will not run a density function, and you will not need a named Bayes formula as a topic.

When every outcome in a finite sample space is equally likely,

P(E) = (number of favorable outcomes) / (number of possible outcomes)

The denominator is a counting problem. The numerator is a counting problem with extra restrictions. Sections 10.1 and 10.2 are the engine; this section names the ratio.

A probability is between 0 and 1 inclusive. P = 0 is impossible. P = 1 is certain. Complements, which 10.4 leans on, use P(not E) = 1 − P(E) because E and not-E partition the sample space.

Worked example. Fair six-sided die. P(rolling a 4) = 1/6. P(even) = 3/6 = 1/2. P(greater than 4) = 2/6 = 1/3. P(at most 2) = 2/6 = 1/3.

Worked example. A bag holds 5 red and 3 blue marbles, well mixed, one marble drawn. P(red) = 5/8. P(blue) = 3/8. These add to 1 because every marble is red or blue.

Worked example that uses combinations. Two cards from 5 distinct cards {A, B, C, D, E}, unordered, equally likely hands. P(the hand includes A) = C(4, 1) / C(5, 2) = 4 / 10 = 2/5. The complementary count is C(4, 2) / 10 = 6/10 = 3/5 for hands that omit A, and 2/5 + 3/5 = 1.

And: Multiply, Then Ask Whether the Second Factor Changes

P(A and B) = P(A) × P(B | A)

P(B | A) is the probability of B given that A has already happened. If A changes the sample space (cards drawn and not replaced, marbles removed), P(B | A) is not P(B).

If A does not change B — separate dice, coins, with-replacement draws, independent machines — then P(B | A) = P(B), and

P(A and B) = P(A) × P(B) for independent events.

Worked example, independent. Two fair coins. P(two heads) = (1/2) × (1/2) = 1/4. Four equally likely ordered outcomes HH, HT, TH, TT; one favorable.

Worked example, independent. Fair die and a fair coin. P(die shows 6 and coin shows heads) = (1/6) × (1/2) = 1/12.

Worked example, dependent. Bag of 5 red and 3 blue. Two marbles without replacement. P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

After the first red is gone, 4 red remain and 7 marbles remain. Using (5/8) × (5/8) = 25/64 would be the with-replacement (or independent) version, and it is a standard wrong answer.

Check with combinations: C(5, 2) / C(8, 2) = 10 / 28 = 5/14. Same fraction. Unordered hands and ordered sequential products agree when you are consistent about both the numerator and the denominator.

Worked example, dependent mixed colors. Same bag, P(red then blue) = (5/8) × (3/7) = 15/56. P(blue then red) = (3/8) × (5/7) = 15/56. P(one of each, in either order) = 15/56 + 15/56 = 30/56 = 15/28.

If the stem says "one red and one blue" without order, add both ordered paths, or use C(5, 1) × C(3, 1) / C(8, 2) = 15 / 28.

With replacement restores independence

Put the marble back. P(both red) = (5/8) × (5/8) = 25/64. The second draw does not know about the first. GMAT signals this with "replaced," "restored," "independent tosses," or separate physical objects (two dice).

Or: Add, Then Subtract the Overlap

P(A or B) = P(A) + P(B) − P(A and B)

The subtraction removes outcomes that sat in both lists and got counted twice. If A and B cannot happen together, they are mutually exclusive (disjoint), P(A and B) = 0, and the formula collapses to P(A) + P(B).

Worked example, mutually exclusive. Fair die. P(2 or 5) = 1/6 + 1/6 = 1/3. A single roll cannot be both 2 and 5.

Worked example, overlap. Fair die. P(even or greater than 4).

• Even: 2, 4, 6 → 3/6
• Greater than 4: 5, 6 → 2/6
• Both (even and greater than 4): 6 → 1/6
• Union: 3/6 + 2/6 − 1/6 = 4/6 = 2/3

List check: {2, 4, 5, 6} has 4 of 6 faces. Adding 3/6 + 2/6 without subtracting produces 5/6, which double-counts the 6.

Worked example, cards. From a 52-card deck, one card. P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. The king of hearts is both a king and a heart.

Worked example, independent events that still overlap. P(A) = 1/3, P(B) = 1/4, A and B independent.

P(A and B) = (1/3)(1/4) = 1/12 P(A or B) = 1/3 + 1/4 − 1/12 = 4/12 + 3/12 − 1/12 = 6/12 = 1/2

Independence does not mean disjoint. Independent events usually can happen together; disjoint events with positive probability are as dependent as events get (if A happened, B cannot have).

Building the Denominator Carefully

Count what the experiment actually is.

• Two dice, ordered pairs: 6 × 6 = 36 outcomes, not 12 and not 6.
• Three coins: 2³ = 8 outcomes.
• Two marbles without replacement, order recorded: 8 × 7 sequential slots.
• Two marbles without replacement, unordered: C(8, 2) = 28 hands.

Mixing an ordered numerator with an unordered denominator (or the reverse) is how 5/14 becomes 20/28 without reducing, or 10/56 left unreduced next to a reduced trap. Pick one convention and stay there.

Worked example. Two fair dice. P(sum is 7). Ordered pairs that sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 pairs. P = 6/36 = 1/6. P(sum is 2) = 1/36 (only 1+1). P(sum is 12) = 1/36. The distribution of sums is not uniform; that is why you count ordered pairs, not the 11 possible sum-values from 2 through 12 as if they were equally likely.

P(sum is 8) uses 5 pairs: (2,6), (3,5), (4,4), (5,3), (6,2), so 5/36. Notice (4,4) appears once as an ordered pair. There is no second (4,4).

Traps on Quant

• Adding P(A) + P(B) when A and B overlap.
• Inventing a fake overlap when A and B are disjoint.
• Using P(A)P(B) without replacement.
• Treating "A or B" as exclusive English "one or the other but not both" when the stem includes the overlap.
• Writing 36 + 36 = 72 for two dice instead of 6 × 6 = 36.
• A probability greater than 1: a sure sign you added when you should have used a union formula, or counted ordered and unordered in the same fraction.
• Replacing P(B | A) with P(A | B) by reading the stem backward.