From a deck, P(King | face card drawn) — there are 12 face cards (J, Q, K in 4 suits): — 1/3.Given a face card was drawn, there are 4 kings among 12 face cards: P = 4/12 = 1/3.
P(A∩B) = 0.15, P(A) = 0.5, P(B) = 0.4. Are A and B independent? — No, because 0.5 × 0.4 = 0.20 ≠ 0.15.P(A)·P(B) = 0.2 ≠ 0.15 = P(A∩B). Since the product rule for independence fails, A and B are dependent.
A die is rolled. P(even | greater than 3): — 2/3.Given >3: outcomes are {4, 5, 6}. Even outcomes: {4, 6} — 2 out of 3. P = 2/3.
P(B|A) · P(A) is equivalent to: — P(A∩B).P(B|A) = P(A∩B)/P(A), so P(B|A)·P(A) = P(A∩B). This is the multiplication rule.
P(Aᶜ|B) = ? — 1 − P(A|B).Given B, P(Aᶜ|B) + P(A|B) = 1 (complementary within the restricted sample space).
How many ways can a president and vice-president be chosen from 10 candidates? — 90.Order matters (two distinct roles): P(10, 2) = 10!/8! = 10 × 9 = 90.
How many ways to choose a 5-card hand from 52 cards? — C(52,5) = 2598960.A hand is unordered (no positions), so use combinations: C(52, 5) = 2,598,960.
C(n, r) = C(n, n−r). Why? — Choosing r to include is equivalent to choosing n−r to exclude.Selecting r items to include is the same as selecting the n−r items to exclude. So C(n,r) = C(n,n−r).
In how many ways can 4 students be arranged in a line from a class of 10? — P(10,4) = 5040.Order matters (positions in line are distinct): P(10, 4) = 10!/6! = 10·9·8·7 = 5040.
How many 3-person committees from 9 people? — 84.Committees are unordered: C(9, 3) = 9!/(3!·6!) = (9·8·7)/(3·2·1) = 504/6 = 84.
The relationship between P(n,r) and C(n,r) is: — P(n,r) = C(n,r)·r!.P(n,r) counts ordered arrangements; C(n,r) counts unordered. P(n,r) = C(n,r) × r! (multiply by the orderings of the r chosen items).
How many distinct 4-letter codes can be made from letters A, B, C, D, E (no repeats)? — 120.Ordered, no repeats: P(5, 4) = 5!/1! = 120.
A pizza shop has 10 toppings. How many different 3-topping pizzas? — C(10,3) = 120.Toppings on a pizza have no order: C(10, 3) = 10!/(3!·7!) = 120.
The number of ways to arrange all 5 letters of the word SMILE: — 5! = 120.Arranging all 5 distinct letters: 5! = 120.
A lottery ticket costs $5. You win $100 with P = 0.03. E(X) = ? — −$2.00.Net: win $95 with P = 0.03 or lose $5 with P = 0.97. E = 95(0.03) + (−5)(0.97) = 2.85 − 4.85 = −$2.00.
X has values {0, 1, 2} with probabilities {0.5, 0.3, 0.2}. E(X) = ? — 0.7.E(X) = 0(0.5) + 1(0.3) + 2(0.2) = 0 + 0.3 + 0.4 = 0.7.
Prior probability P(A) is called the: — Prior.P(A) before observing any evidence is the "prior probability." After updating with evidence, we get the posterior.
After applying Bayes' Theorem, P(A|B) is called the: — Posterior.P(A|B) — the updated probability after incorporating evidence B — is called the posterior probability.
If E(X) = 5 and you play 100 times, expected total winnings ≈ ? — $500.Expected total = n · E(X) = 100 · 5 = $500. This uses the linearity of expectation.
Bayes' Theorem is especially useful when: — Updating probabilities with new evidence.Bayes' Theorem is the tool for updating prior beliefs when new evidence arrives — the foundation of Bayesian inference.
A raffle has 1000 tickets at $2 each, with 1 prize of $500. E(net) per ticket = ? — $−1.50.P(win) = 1/1000. E(winnings) = 500(1/1000) + 0(999/1000) = $0.50. E(net) = 0.50 − 2.00 = −$1.50.
You win $3 on heads (P=0.5) and lose $2 on tails (P=0.5). After 10 flips, expected net = ? — $5.E per flip = 3(0.5) − 2(0.5) = 0.5. After 10 flips: 10 × 0.5 = $5.