MATH U113 · Doubt clinic · topic page
Sets, complements and the addition rule
Events as sets, De Morgan, the four regions of a Venn diagram, and the one addition rule (with its overlap term zero when events cannot co-occur). Notes: §2.1, §2.2. Cards are collapsed — open one, or use the buttons. Newest at the bottom. Back to the clinic index.
They answer different questions. Combinations count outcomes: C(n, k) is a number of cases, and it goes on top or bottom of "favourable over total" — legal only when every selection is equally likely (a shuffled deck, a random order, "at random"). The addition rule combines probabilities of events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), for events whose probabilities you already have or have just computed. Ask, in this order:
- Are the inputs counts of objects, or probabilities of events? Counts (52 cards, 15 phones, 12 widgets) → combinations. Probabilities (P(A) = ½, "works with probability 0.9") → addition / multiplication rules.
- Is "or" joining two events, or describing one selection? "King or ace" joins two events → addition rule (overlap zero here). "Two aces and three jacks" describes one selection → a product of two combination counts. "At least one …" → usually 1 − a single count.
- Can you draw the sample space? If it is small (36 dice cells, 24 orders), draw it: the addition rule becomes "count A, count B, remove the cells counted twice", and a combination is just the size of a region.
When both appear in one problem the order is fixed: combinations first to get each event's probability, then the addition rule to join events that cannot overlap (Q8(b): count inside each case, add the three cases). Two traps: adding where the question says "and" (that multiplies), and using a combination when the cases are not equally likely (a biased component, boxes of different sizes — Q13). One line for the exam: counts of objects go into C(n, k); probabilities of events go into the addition rule; "or" adds and subtracts the overlap, "and" multiplies.
Lecture slide 17 · 13 SepA king or an aceOne card from a shuffled deck: P(king or ace)? Why no overlap term here but one in the dice example.
Where this lives Notes §2.2 · addition rule, equally likely outcomes · Lesson §2 · the rules · Devore §2.2 · kit: a deck has 4 suits × 13 ranks
On the board this one was written P(A ∨ B) = P(A) + P(B), and the very next example (card 2) subtracts something. Two formulas for "or"? No — one rule, and in this problem its last term happens to be zero. Also, notation: the lecturer's ∨ is "or" = ∪, and ∧ is "and" = ∩; Devore and the notes use the set symbols.
Step 1Name the sample space and the counting rule you're allowed to use
One card from a well-shuffled deck: the sample space is the 52 cards, and "well shuffled" is the phrase that makes them equally likely. That licenses the school formula (notes §2.2, last bullet): P(E) = N(E)/52. Every probability in this problem is a count divided by 52.
Step 2Name the events and ask the one question that decides the formula
A = "the card is a king": 4 cards (one per suit). B = "the card is an ace": 4 cards. Now the deciding question: can one card be in both? A single card has one rank; it cannot be a king and an ace. So A ∩ B = ∅ — the events are mutually exclusive (notes §2.1) — and P(A ∩ B) = 0.
Step 3Apply the one addition rule
The slide simply skipped writing the "− 0". Writing it anyway is good exam hygiene: it shows you checked the overlap rather than assumed it away. Sanity: 8 favourable cards out of 52, and 8/52 = 2/13 ≈ 0.154 ✓.
1) Adding without checking the overlap. "King or heart" adds 4/52 + 13/52 and then must remove the king of hearts (twin below). The addition-without-subtraction form is only for mutually exclusive events. 2) "Or" is not "and". "King or ace" is a union — 8 cards. "King and ace" is impossible for one card — probability 0. 3) Multiplying. 4/52 × 4/52 answers a different question (two draws with replacement, king then ace) — "or" never multiplies.
Try the twin: one card is drawn. Find the probability it is a king or a heart.
A = king: 4/52. B = heart: 13/52. Overlap: the king of hearts is both, so P(A ∩ B) = 1/52 — not zero this time. P(A ∪ B) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. Check by counting: 13 hearts plus the 3 kings that are not hearts = 16 cards ✓.
Lecture slide 17 · 13 SepEven first die or total 8Two dice: P(even on the first die or a total of 8)? Where 18, 5 and 3 come from and why 3 is subtracted.
Where this lives Notes §2.2 · addition rule (the Venn figure there is this card's Step 3) · Lesson §2 · Devore §2.2 · kit: listing ordered pairs
Three numbers appear from nowhere on the slide — 18, 5, 3 — and then one of them is subtracted. This card shows where each comes from by drawing the sample space once, and then why the subtraction is forced: without it, three outcomes get counted twice.
Step 1Draw the sample space — ordered pairs, 36 of them
The board listed S = {(1,1), (1,2), …, (6,6)}. Each outcome is a pair (first die, second die), and the order matters: (2, 6) and (6, 2) are different outcomes (a red 2 and a blue 6 is not the same as a red 6 and a blue 2). Six choices for the first × six for the second = 36 equally likely outcomes — the "equally likely" again coming from fair dice. So every probability here is a count over 36.
Step 2Count each event on the grid
- A = even number on the first die. The first die shows 2, 4 or 6 — three rows of the grid — and the second die can be anything: 3 × 6 = 18 outcomes.
- B = total of 8. List them: first die 2 needs second 6; 3 needs 5; 4 needs 4; 5 needs 3; 6 needs 2 (first die 1 would need a 7 — impossible). B = {(2,6), (3,5), (4,4), (5,3), (6,2)}: 5 outcomes, a diagonal on the grid.
- A ∩ B = both. Which of those five have an even first die? (2,6), (4,4), (6,2): 3 outcomes.
Step 3Add, then remove the double count
Look at the dark cells. Each is in the 18 (its row is even) and in the 5 (it sums to 8). So 18 + 5 = 23 counts each of them twice; the union really has 18 + 5 − 3 = 20 outcomes. That is all the addition rule says:
Cross-check without the formula (do this in the exam, it takes ten seconds): count the union directly on the grid — the 18 shaded cells, plus the outlined cells not already shaded, (3,5) and (5,3): 18 + 2 = 20 ✓. If your formula answer and your direct count disagree, the overlap count is wrong.
1) 23/36. Adding and forgetting the overlap — the single most common slip in §2.2. Rule of thumb: if the two events can happen together, there is an overlap to subtract. 2) Unordered dice. Treating (2,6) and (6,2) as one outcome gives 21 "outcomes" that are not equally likely, and every count breaks. Keep the pairs ordered. 3) Mixing "first die even" with "at least one die even". The latter is 27 outcomes, not 18 — read which die the question names.
Try the twin: two dice are tossed. Find the probability of an odd number on the first die or a total of 7.
A = odd first die: rows 1, 3, 5 → 18. B = total 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6. Overlap (odd first die among those): (1,6), (3,4), (5,2) → 3. P = 18/36 + 6/36 − 3/36 = 21/36 = 7/12. Direct count: 18 + the three B-cells with even first die = 21 ✓.
Problem set Q3 · 14 SepP(A ∪ Bᶜ) from three numbersP(A) = P(B) = ½, P(Aᶜ ∩ Bᶜ) = ⅓: find P(A ∪ Bᶜ) by filling the four Venn regions.
Where this lives Notes §2.1 · De Morgan · Notes §2.2 · addition rule · Quiz post-mortem Q2 (same Venn) · Devore §2.2
Nothing says A and B are independent, so you cannot multiply anything. The only tool is the Venn diagram's four regions: fill in what the data pin down, then read off the target. Three given numbers, four regions that sum to 1 — exactly enough.
Step 1Translate the third datum with De Morgan
Ac ∩ Bc = (A ∪ B)c — "neither" is the outside of the union. So P(A ∪ B) = 1 − ⅓ = ⅔.
Step 2Addition rule gives the overlap
Now all four regions are known: A only = ½ − ⅓ = ⅙; both = ⅓; B only = ⅙; neither = ⅓. Check: ⅙ + ⅓ + ⅙ + ⅓ = 1 ✓.
Step 3Read the target off the diagram
A ∪ Bc = "in A, or outside B" — its complement is the single region "B only" = Ac ∩ B, with probability P(B) − P(A ∩ B) = ½ − ⅓ = ⅙. So
Or add the three regions directly: ⅙ + ⅓ + ⅓ = ⅚ ✓.
1) Assuming independence. P(A ∩ B) = ¼ is wrong here; the data force ⅓. 2) Applying the addition rule to A and Bc without P(A ∩ Bc). It works — P(A) + P(Bc) − P(A only) = ½ + ½ − ⅙ = ⅚ — but only once you have the regions; the complement route is one step shorter. 3) Reading Ac ∩ Bc as (A ∩ B)c. De Morgan: "neither" is the complement of the union.
Try the twin: P(A) = 0.6, P(B) = 0.5, P(Aᶜ ∩ Bᶜ) = 0.2. Find P(Aᶜ ∪ B).
P(A ∪ B) = 0.8; overlap = 0.6 + 0.5 − 0.8 = 0.3; regions: A only 0.3, both 0.3, B only 0.2, neither 0.2. Ac ∪ B is everything except "A only": 1 − 0.3 = 0.7.