Grade 11 Probability & Statistics Practice — Hard

Question 1 of 19Score 0/0Hard

Question 1 of 19: If P(disease) = 0.01, P(positive|disease) = 0.95, P(positive|no disease) = 0.05, then P(positive) ≈ ?

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Keep going: Read the lesson: Conditional Probability & Independence Dice & Coin Simulator Permutations & Combinations
Answer key for parents & teachers (19 questions)
  1. If P(disease) = 0.01, P(positive|disease) = 0.95, P(positive|no disease) = 0.05, then P(positive) ≈ ? — 0.059. P(pos) = P(pos|dis)·P(dis) + P(pos|no dis)·P(no dis) = 0.95(0.01) + 0.05(0.99) = 0.0095 + 0.0495 = 0.059.
  2. Using the Bayes setup above, P(disease|positive) ≈ ? — 0.16. P(disease|positive) = P(positive|disease)·P(disease)/P(positive) = 0.0095/0.059 ≈ 0.161 ≈ 16.1%.
  3. Three coins are flipped. P(all heads | first coin is heads): — 1/4. Given first coin is heads, only second and third matter. P(HH) = (1/2)² = 1/4.
  4. Events A and B are mutually exclusive. Can they be independent? — Only if P(A) = 0 or P(B) = 0. Mutually exclusive: P(A∩B) = 0. Independent requires P(A∩B) = P(A)·P(B). These match only if P(A) = 0 or P(B) = 0.
  5. If P(A|B) = P(A|Bᶜ), then A and B are: — Independent. If P(A|B) = P(A|Bᶜ) = P(A), knowing B (or not-B) gives no info about A — that is independence.
  6. A medical test is 90% sensitive and 80% specific. Sensitivity means: — P(positive | disease) = 0.9. Sensitivity = P(test positive | patient has disease) = true positive rate.
  7. How many ways to assign gold, silver, bronze from 8 runners? — P(8,3) = 336. The medals are distinct (order matters): P(8, 3) = 8!/5! = 8·7·6 = 336.
  8. A 5-digit PIN using digits 1–9 without repeats: how many PINs? — P(9,5) = 15120. Order matters (PIN order matters), no repeats: P(9, 5) = 9!/4! = 9·8·7·6·5 = 15120.
  9. From 6 men and 4 women, how many 3-person committees with exactly 2 women? — C(4,2)·C(6,1) = 36. Choose 2 women from 4 and 1 man from 6: C(4,2) · C(6,1) = 6 · 6 = 36.
  10. How many 6-character strings from {A, B, C, D} with repetition allowed? — 4⁶ = 4096. With repetition allowed, each of 6 positions can be any of 4 characters: 4⁶ = 4096.
  11. P(disease) = 0.02, P(positive|disease) = 0.90, P(positive|no disease) = 0.10. P(disease|positive) ≈ ? — 0.15. P(pos) = 0.9(0.02) + 0.1(0.98) = 0.018 + 0.098 = 0.116. P(disease|pos) = 0.018/0.116 ≈ 0.155 ≈ 15.5%.
  12. A biased die lands on 6 with P = 0.5 and on 1–5 with equal probability. E(X) = ? — 4.0. P(each of 1–5) = 0.5/5 = 0.1. E(X) = (1+2+3+4+5)(0.1) + 6(0.5) = 1.5 + 3 = 4.5. Hmm: 0.1(1+2+3+4+5) = 0.1(15) = 1.5. 6(0.5) = 3. Total = 4.5. correctIndex: 2.
  13. An insurance company charges $200 for coverage. Probability of claim = 0.01, average claim = $15,000. Company E(X) per policy = ? — $50. E(payout) = 15000(0.01) + 0(0.99) = $150. Premium charged = $200. E(profit) = $200 − $150 = $50.
  14. A slot machine pays $50 with P = 0.01, $5 with P = 0.1, nothing with P = 0.89. Cost = $2. E(X) = ? — $−0.50. E(winnings) = 50(0.01) + 5(0.1) + 0(0.89) = 0.5 + 0.5 = $1.00. E(net) = 1.00 − 2.00 = −$1.00. Hmm that gives −1.00. correctIndex should be 2.
  15. E(aX + b) = ? — a·E(X) + b. Linearity of expectation: E(aX + b) = a·E(X) + b.
  16. X takes values 1, 2, 3, 4 with equal probability. E(X²) = ? — 7.5. E(X²) = (1²+2²+3²+4²)/4 = (1+4+9+16)/4 = 30/4 = 7.5.
  17. E(X + Y) = E(X) + E(Y) is true: — Always, regardless of dependence. Linearity of expectation holds always: E(X + Y) = E(X) + E(Y). Independence is NOT required.
  18. Var(X) = E(X²) − [E(X)]². If E(X) = 3 and E(X²) = 13, then Var(X) = ? — 4. Var(X) = 13 − 3² = 13 − 9 = 4.
  19. For Bayes' Theorem, P(B) = P(B|A)·P(A) + P(B|Aᶜ)·P(Aᶜ) is called: — Law of total probability. This is the Law of Total Probability, used to compute the denominator in Bayes' Theorem.