MAT-144 · Mathematical Reasoning Topic 06 · Probability
Lesson 03 · OR · the addition rule

Sample spaces and the addition rule.

P(A or B) = P(A) + P(B) − P(A and B). The standard deck of 52 cards is the canonical playground.

01Build the sample space 02P(A) + P(B) 03Subtract P(A and B)
▸ THE HOOK

Draw one card from a shuffled deck. What's the probability the card is red OR a face card? You know P(red) = 26/52 = 1/2 from L1 (half the deck is red). You know P(face) = 12/52 because there are 12 face cards (J, Q, K in each of four suits). So is P(red OR face) = 26/52 + 12/52 = 38/52?

No. That answer double-counts the six cards that are both red AND a face: J♥, Q♥, K♥, J♦, Q♦, K♦. Each one gets counted once when you tally "red" and a second time when you tally "face." To get the right count, you have to subtract the overlap once.

That correction — "don't count the overlap twice" — is the addition rule. It's the formula that handles every OR question in probability, and it's where the standard 52-card deck earns its place as the training ground for the topic.

Add the parts; subtract the overlap.

For any two events A and B in the same sample space, the probability that A or B (or both) occurs is given by the addition rule:

P(A or B) = P(A) + P(B) − P(A and B)

The first two terms count every outcome in A and every outcome in B. But outcomes in BOTH — the overlap — get counted by each of the first two terms, so they appear twice in the sum. Subtracting P(A and B) once removes the duplicate, leaving each outcome counted exactly once.

When A and B share no outcomes (the overlap is empty), P(A and B) = 0 and the formula reduces to the simpler P(A) + P(B). Events with no overlap are called mutually exclusive — rolling a 3 and rolling a 5 on the same die, for instance: they can't both happen at once.
P(A OR B) — DON'T DOUBLE-COUNT add the two pieces, then subtract the overlap once red 20 cards both 6 cards face 6 cards A = RED 26/52 B = FACE 12/52 P(A or B) = 26/52 + 12/52 − 6/52 = 32/52 = 8/13 six face cards are also red — subtracting prevents counting them twice

The deck split two ways. Red contains 26 cards; Face contains 12 cards; the intersection (red AND face) contains 6 cards — the J/Q/K of hearts and diamonds. The total "red OR face" region is the two circles' union: 26 + 12 − 6 = 32 cards.

▸ DEFINITION

The addition rule for two events A and B states P(A or B) = P(A) + P(B) − P(A and B). When A and B are mutually exclusive (they cannot both happen), the overlap term is zero and the rule simplifies to P(A or B) = P(A) + P(B).

Words you'll see on ALEKS

  • Union (A or B) The event "A happens OR B happens (or both)." Includes every outcome that's in either set. In set notation: A ∪ B.
  • Intersection (A and B) The event "A happens AND B happens at the same time." Includes only the overlap. In set notation: A ∩ B. The intersection is what the addition rule subtracts.
  • Mutually exclusive Two events are mutually exclusive when they cannot both occur on the same trial. Rolling a 3 and rolling a 5 on the same die roll. Their intersection is empty, so P(A and B) = 0 and the addition rule simplifies.
  • Standard 52-card deck 52 cards in 4 suits (♠ ♣ ♥ ♦), 13 cards per suit (A, 2-10, J, Q, K). Hearts and diamonds are red (26 cards); spades and clubs are black (26 cards). Face cards are J, Q, K (12 total, three per suit). The canonical deck shows up in every ALEKS probability review.
  • Double-counting The mistake of including the same outcome twice. Happens when you compute P(A or B) as P(A) + P(B) without subtracting the overlap. Always subtract P(A and B) — the inclusion-exclusion rule is the fix.

P(face), P(red), P(face or red).

Direct from ALEKS Q3. One card is drawn from a standard 52-card deck. Three parts, increasing in complexity: a simple count, another simple count, and a union that requires the overlap correction.

"Suppose one card is drawn at random from a standard deck of 52 cards. Find: (a) P(face), (b) P(red), (c) P(face OR red). Write your answers as fractions."

1

(a) P(face) — count the face cards.

Face cards are J, Q, K in each of four suits: 3 × 4 = 12 face cards out of 52 total.

P(face) = 12/52 = 3/13
→ 12/52 reduces (GCD = 4)
2

(b) P(red) — count the red cards.

Two of the four suits are red (hearts and diamonds): 2 × 13 = 26 red cards.

P(red) = 26/52 = 1/2
→ exactly half the deck
3

(c) P(face OR red) — addition rule.

Apply the formula, but first identify the overlap. The cards that are both face AND red are the J/Q/K of hearts and diamonds: 6 cards. So P(face AND red) = 6/52.

P(face or red) = P(face) + P(red) − P(face and red)
= 12/52 + 26/52 − 6/52
= 32/52 = 8/13

Sanity check: the union has 32 cards total — 26 red cards plus 6 black face cards (J♠, Q♠, K♠, J♣, Q♣, K♣) that are face but not red. 32/52 = 8/13.

→ subtract the overlap once

Three OR questions.

Same rule, different scenarios. Identify whether the overlap is zero (mutually exclusive) or non-zero (need to subtract). Write every answer as a fraction in lowest terms.
PROBLEM 01 ☆ ☆   warm-up · die roll (mutually exclusive)

A fair six-sided die is rolled. What is P(roll a 2 OR roll a 5)?

P =
PROBLEM 02 ★ ★ ☆   deck draw with overlap

One card is drawn from a standard 52-card deck. What is P(king OR heart)?

P =
PROBLEM 03 ★ ★ ★   students with overlap

Of 30 students, 18 play soccer, 12 play basketball, and 7 play both. One student is picked at random. What is P(soccer OR basketball)?

P =

Three fast questions before you move on.

Tap an answer. Feedback shows up immediately.

Q1. Two events A and B are mutually exclusive. Which formula computes P(A or B)?

Why B? Mutually exclusive means the events can't both happen, so P(A AND B) = 0. The general addition rule P(A) + P(B) − P(A and B) collapses to just P(A) + P(B). Option C is technically correct too (it simplifies to B when the overlap is zero), but B is the canonical mutually-exclusive form.

Q2. A card is drawn from a standard deck. What is P(spade or club)?

Why B? Spades and clubs are different suits — mutually exclusive. P(spade) = 13/52, P(club) = 13/52, no overlap. P = 13/52 + 13/52 = 26/52 = 1/2 (the two black suits).

Q3. If P(A) = 0.4, P(B) = 0.5, and P(A AND B) = 0.2, what is P(A OR B)?

Why C? P(A) + P(B) − P(A and B) = 0.4 + 0.5 − 0.2 = 0.7. A is the double-counting trap (0.4 + 0.5 = 0.9 without subtracting the overlap). D is what you'd get if you added the overlap instead of subtracting (which would make the answer greater than 1, impossible).
▸ UP NEXT — LESSON 04

OR questions, anywhere they appear.

The addition rule shows up every time you want to combine two events with OR. Polls ("voters who lean left OR are under 30"), epidemiology ("patients with diabetes OR high blood pressure"), insurance ("claim filed for collision OR theft") — the structure is identical, just larger sample spaces.

The single most important habit: before you add, ask whether the overlap is zero. If the events are mutually exclusive, you can skip the subtraction. If they're not, subtract the intersection once to avoid double-counting.

Next: Lesson 4 introduces permutations and combinations — what to do when the counting principle's "independent steps" assumption breaks down because each choice removes an option from the next step.

Continue to Lesson 04

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