MATH U113 · Doubt clinic · topic page
Counting with equally likely outcomes
Favourable over total, in the same mode top and bottom: hands, positions in a random order, derangements, inclusion–exclusion. Notes: §2.3; drill: counting drill. 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 · 13 Sep2 aces and 3 jacks in a poker handFive-card hand: P(2 aces and 3 jacks)? Why combinations, why multiply, why C(52,5) underneath.
Where this lives Notes §2.3 · combinations, the "same mode" rule, worked example 1 (the defective-boards example is this exact pattern) · Lesson §3 · counting from zero · Devore §2.3 · drill: counting drill
Four questions hide in one board line: why is it C (combinations) and not something with order? why multiply the two C's? why is 52C5 underneath? and why is there no other factor for "the rest of the hand"? Each gets one step below.
Step 0What a "hand" is — and why that decides everything
A poker hand is 5 cards dealt from 52. You hold them as a set: getting A♠ then J♥ then … is the same hand as getting them in any other order. So a hand is an unordered selection of 5 from 52 — a combination, written 52C5 or (525). Well shuffled ⇒ every one of those hands is equally likely, so
The golden rule from notes §2.3: top and bottom must count in the same mode. We chose unordered for the bottom, so the top must be unordered too.
Step 1The denominator: all hands
Unpack this step: why five falling factors over 5!
Dealing 5 cards in order has 52 · 51 · 50 · 49 · 48 outcomes. Each unordered hand appears among them once per ordering of its 5 cards, i.e. 5! = 120 times. Divide the overcount out (notes §2.3, "why the subset formula divides by k!").
Step 2The numerator: build one favourable hand, count the choices
To hold 2 aces and 3 jacks you must make two decisions, and the hand is then complete:
- Which 2 of the 4 aces? Order irrelevant (it's a hand) → 4C2 = 4·3/2! = 6 ways.
- Which 3 of the 4 jacks? → 4C3 = 4·3·2/3! = 4 ways (equivalently, choose which jack to leave out: 4).
Why multiply: every ace-pair can be combined with every jack-triple, and each combination is a different hand — the product rule (notes §2.3, first row of the table). The grid below is that multiplication: 6 rows × 4 columns = 24 hands.
Why nothing else: 2 + 3 = 5. The hand is full; there is no "remaining cards" factor to choose. (Contrast the twin, where there is.)
Step 3Divide
Sanity: a specific full house is rare — about one hand in a hundred thousand feels right. (All full houses together are 3744 hands, ≈ 0.14%; this is one of the 156 rank-patterns among them, and 3744/156 = 24 ✓.)
Unpack this step: the ordered route gives the same number
Deal the cards in order. One specific sequence, say A A J J J, has probability (4/52)(3/51)(4/50)(3/49)(2/48); the two aces can occupy any 2 of the 5 positions, 5C2 = 10 patterns, each with the same probability. 10 × 288/311 875 200 = 2880/311 875 200 = 1/108 290 ✓ — same answer, because ordered/ordered is also "same mode". Combinations are just less work.
1) Mixed modes. Permutations on top with 52C5 underneath (or the reverse) gives an answer off by a factor of 5! or 120 — sometimes bigger than 1, which is your alarm bell. 2) Multiplying by 5! "for the arrangements". The hand is a set; you already chose unordered. 3) Forgetting the rest of the hand. When the specification does not fill all 5 cards — "exactly 2 aces" — the other cards must be chosen too, from the non-aces (twin below). Count the cards named; if they total 5, stop. 4) 4C2 + 4C3. Adding counts either an ace-pair or a jack-triple; a hand needs both, and "both" multiplies.
Try the twin: find the probability that a 5-card hand contains exactly 2 aces.
Choose the 2 aces: 4C2 = 6. The hand still needs 3 more cards, and they must be non-aces (otherwise it would be 3 or 4 aces): choose 3 from the 48 non-aces, 48C3 = 48·47·46/6 = 17 296. Favourable hands: 6 × 17 296 = 103 776. P = 103 776 / 2 598 960 ≈ 0.0399, about 4%. Notice the extra factor that card 3 didn't need — because "2 aces and 3 jacks" already filled the hand.
Problem set Q5 · 14 SepThree hats, nobody gets his ownThree men take hats at random: P(no one gets his own) — a derangement, listed.
Where this lives Module 2 · the matching problem · Notes §2.3 · derangements line · Quiz post-mortem Q3 (the same problem with four PINs)
This is the quiz's Q3 in a smaller costume: "none gets his own" = a derangement. With three items, list all six orders and count — no formula needed.
Step 1List the 3! = 6 equally likely assignments
Write the hats the men 1, 2, 3 receive in order. A man is "matched" when position i holds hat i.
Step 2Count and divide
Bonus, for the pattern: P(exactly one matched) = 3/6, P(all three) = 1/6, P(exactly two) = 0 — so E(matches) = (3 + 3)/6 = 1, the matching-problem fact from the quiz.
1) (2/3)³ = 8/27. "Each man has probability 2/3 of a wrong hat" is true, but the three events are not independent (if two men have wrong hats, the third's is forced), so you cannot multiply. 2) Complement = "all get their own". "None" is not the complement of "all"; the complement of "none" is "at least one".
Try the twin: four men. Probability nobody gets his own hat?
D4/4! = 9/24 = 3/8 = 0.375 — the 9-out-of-24 count from the quiz post-mortem's permutation grid.
Problem set Q7 · 14 SepSix cereal packets, three favourite dinosaursFive equally likely models per packet: P(all three favourites appear in six packets) — inclusion–exclusion on "missing".
Where this lives Notes §2.2 · addition rule for three events · Notes §2.5 · independence · Devore §2.2, 2.5
"Acquires all three favourites" is an and of three things happening somewhere in six packets — hard to count directly. The move is to flip to the complement: "at least one favourite is missing", and that is an or of three events, which the three-event addition rule (inclusion–exclusion) handles. The packets are independent, so "favourite i missing from all six" is a power.
Step 1Name the bad events
M1, M2, M3 = "favourite 1 (2, 3) never appears in the six packets". Wanted: P(none missing) = 1 − P(M1 ∪ M2 ∪ M3).
Step 2Probability of each bad event and each overlap
One packet avoids favourite 1 with probability 4/5; six independent packets: P(M1) = (4/5)6. Avoiding two named favourites: 3/5 per packet, P(M1 ∩ M2) = (3/5)6. All three: (2/5)6. By symmetry every single, pair and triple has the same value.
Step 3Inclusion–exclusion, then complement
Unpack this step: the powers
4⁶ = 4096, 3⁶ = 729, 2⁶ = 64, 5⁶ = 15625. 3·4096 = 12288, 3·729 = 2187.
Sanity: about a one-in-three chance to collect three specific dinosaurs in six packets feels right — six draws from five types usually miss at least one.
1) (1/5)³. That is "the first three packets are favourites 1, 2, 3 in order" — a completely different event. 2) Stopping at 3(4/5)⁶. The three "missing" events overlap; adding them double-counts the pairs. Subtract pairs, add back the triple — the three-event rule in notes §2.2. 3) "Favourite" ≠ "any three". The question fixes which three dinosaurs; nothing is chosen.
Try the twin: four packets, two favourite dinosaurs. Probability both favourites appear?
1 − [2(4/5)⁴ − (3/5)⁴] = 1 − (512 − 81)/625 = 194/625 = 0.3104.
Problem set Q8 · 14 SepFifteen phones in a random service orderAll cordless in the first ten; only two types left after ten (the hint decoded); two of each type in the first six.
Where this lives Notes §2.3 · combinations, same-mode rule, worked example 1 · Doubt card D3 (poker hand — same shape) · Devore §2.3
The random numbering is a red herring for counting: what matters is which set of phones lands in the first ten positions (or first six). Every 10-subset of the 15 phones is equally likely to be "the first ten", so all three parts are favourable subsets over C(15, 10) = C(15, 5) = 3003 — the poker-hand pattern. Part (b) adds one wrinkle: an "either–or" that must not be double-counted.
Step 1(a) All five cordless among the first ten
Think of the last five positions instead: they must contain no cordless phone, i.e. be any 5 of the 10 non-cordless phones. P = C(10, 5)/C(15, 5) = 252/3003 = 12/143 ≈ 0.08392. (Equivalent: choose which 10 phones go first, C(15,10) = 3003; favourable = all 5 cordless plus 5 of the other 10, C(10,5) = 252. Same numbers.)
Step 2(b) Exactly two types remain — the hint, decoded
"Only two types remain after ten" means exactly one type was completely used up in the first ten. Count by naming that type:
- Pick the used-up type: 3 ways. Say cordless.
- All five cordless are among the first ten, plus five of the other ten phones: C(5,5) · C(10,5) = 252 ways.
- But two of those 252 choices use up a second type as well (the other five are all cellular, or all corded) — then only one type remains, which is not what we want. Remove them: 252 − 2 = 250.
- Three types could be the used-up one, and these cases cannot overlap (exactly one type is gone in each): 3 × 250 = 750.
Unpack this step: why the "− 2" and why not "− 2" three times over
Within the cordless-used-up case, the other five in the first ten are chosen from 5 cellular + 5 corded. Exactly two of those C(10,5) = 252 choices are "all five cellular" or "all five corded" — each would empty a second type. Subtracting them inside the case, before multiplying by 3, is right; the three cases (which type is gone) are disjoint, so no further correction.
Step 3(c) Two of each type among the first six
First six positions = a 6-subset, C(15,6) = 5005 equally likely. Favourable: 2 of the 5 cellular, 2 of the 5 cordless, 2 of the 5 corded — product rule: C(5,2)3 = 103 = 1000.
1) Ordering the top but not the bottom. If you count arrangements of the first ten (10! ways), the denominator must be 15!/5!; mixing modes is the poker-hand trap again. 2) (b) as "some type is gone" = 3 × 252. That counts the two-types-gone arrangements twice and includes them at all when they are excluded by "only two remain". 3) Reading (a) as "the first ten are all cordless". Impossible — there are only five.
Try the twin: P(no cordless phone among the first six)?
Choose the first six from the 10 non-cordless: C(10,6)/C(15,6) = 210/5005 = 6/143 ≈ 0.04196.
Problem set Q11 · 14 SepWidget inspector: at least 9 tests3 defectives among 12, tested one by one: P(at least 9 tests), P(at least 10) — position of the last defective.
Where this lives Notes §2.3 · combinations, "positions" counting · Notes §2.2 · complement rule · Devore §2.3
"How many tests" sounds like a new kind of random variable, but it is a positions question in disguise: the number of tests is the position of the last defective in the random testing order. "At least 9 tests" just says that position is 9 or more — and its complement, "all three defectives sit in the first 8 positions", is a subset count.
Step 1Model: the three defectives occupy a random 3-subset of the 12 positions
Mixing the widgets makes every choice of 3 positions (out of 12) for the defectives equally likely: C(12, 3) = 220 cases.
Step 2Complement and count
Check: (b) ≤ (a) ✓ ("at least 10" is a sub-case of "at least 9").
Unpack this step: a subtlety the key ignores
A clever inspector who has tested 9 widgets and found them all good knows the last 3 are defective and could stop at 9. The sheet's model is "test until the third defective is physically found"; under that reading the answers are 41/55 and 34/55. Use that reading in the exam, and mention the alternative only if asked.
1) "At least 9" as "exactly 9". Exactly 9 tests means the third defective is at position 9: C(8,2)/C(12,3) = 28/220. Different question. 2) Sequential probabilities. Multiplying 3/12 · 2/11 · … works only if you also enumerate all the orders; the subset count avoids it. 3) Positions of the good widgets. Choosing where the 9 good ones go, C(12,9) = 220, is the same count — fine either way, but don't mix the two.
Try the twin: P(at least 11 tests)?
1 − C(10,3)/C(12,3) = 1 − 120/220 = 100/220 = 5/11 ≈ 0.45455.
Devore Example 2.22 · 15 SepiPod shuffle: the first Beatles song is the fifth played100 songs, 10 by the Beatles, random order: P = (90·89·88·87·10)/(100·99·98·97·96) = .0679 — and why combinations give the same number.
Where this lives Notes §2.3 · permutations, combinations, the same-mode rule · Widget inspector card · Blood typing card (the same event in conditional-probability form) · Devore §2.3
The book solves it twice, with ordered counts and then with combinations, and the two calculations look nothing alike. They are the same fraction in two costumes. The event is concrete: songs 1–4 are non-Beatles (NB) and song 5 is Beatles (B). Count that in whichever mode you like — as long as the denominator is in the same mode.
Step 1Ordered route: the first five songs as a sequence
All ordered choices of the first five songs are equally likely: 100 · 99 · 98 · 97 · 96 sequences. Favourable sequences: an NB for slot 1 (90 choices), an NB for slot 2 (89 left), 88, 87, then a B for slot 5 (10 choices):
Read it as the multiplication rule with a shrinking pool: 90/100, then 89/99, 88/98, 87/97, and finally 10/96 — five fractions multiplied. It is the blood-typing calculation with bigger numbers.
Step 2Unordered route: which positions the Beatles songs occupy
Forget the order of play and ask only which 10 of the 100 positions hold the Beatles songs: C(100, 10) equally likely position-sets. Favourable: position 5 is a B, positions 1–4 are not, and the other 9 Bs sit anywhere in positions 6–100: C(95, 9) ways.
One Beatles song is pinned to position 5. The other nine are not offered positions 1–4 (that would make an earlier Beatles song) and position 5 is taken, so they choose from positions 6–100 only: 95 positions, C(95, 9). Every set counted this way has its first Beatles song at 5. If the first one were allowed anywhere from 1 to 5, the nine others could use any of the 99 remaining positions: C(99, 9)/C(100, 10) = 10/100 = .1 — simply "position 5 holds a Beatles song", first or not. The drop from .1 to .0679 is the cost of demanding four non-Beatles songs in front. The ordered route enforces the same thing slot by slot with its 90, 89, 88, 87.
Unpack this step: why C(95, 9)/C(100, 10) equals the ordered fraction
Expand both: C(95,9)/C(100,10) = [95!/(9! 86!)] · [10! 90!/100!] = 10 · (90 · 89 · 88 · 87)/(100 · 99 · 98 · 97 · 96), after cancelling 95!/100! = 1/(100·99·98·97·96) and 90!/86! = 90·89·88·87, and 10!/9! = 10. Same fraction, so the two answers cannot disagree.
Same-mode rule in action: route 1 counts ordered sequences top and bottom; route 2 counts position-sets top and bottom. Mixing them is the error.
1) (9/10)⁴ · (1/10). That treats each song as independent with replacement; the playlist has no replacement, so the pool shrinks. 2) Forgetting the "first Beatles" part. "The fifth song is a Beatles song" alone is 10/100 = .1; the question also requires the first four to be non-Beatles. 3) 10 · 4 or similar for "which slot". The slot is fixed at 5 by the question; nothing is chosen there.
Try the twin: P(the first Beatles song is the third song played)?
(90 · 89 · 10)/(100 · 99 · 98) = 80100/970200 = .0826. Higher than for the fifth slot, as it should be: an earlier first Beatles song is more likely than a later one when 10% of the list is Beatles.