Grade 11 Probability & Statistics Practice — Easy

Question 1 of 26Score 0/0Easy

Question 1 of 26: P(A|B) is read as:

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Keep going: Read the lesson: Conditional Probability & Independence Dice & Coin Simulator Permutations & Combinations
Answer key for parents & teachers (26 questions)
  1. P(A|B) is read as: — P of A given B. P(A|B) = "the probability of A given that B has occurred." It conditions on B.
  2. The formula for conditional probability is: — P(A∩B)/P(B). P(A|B) = P(A∩B)/P(B). This restricts the sample space to outcomes where B occurred.
  3. Two events A and B are independent if: — P(A∩B) = P(A)·P(B). Independence: P(A∩B) = P(A)·P(B). Knowing B occurred does not change the probability of A.
  4. P(A) = 0.3, P(B) = 0.6, events independent. P(A∩B) = ? — 0.18. Independent: P(A∩B) = P(A)·P(B) = 0.3 × 0.6 = 0.18.
  5. Drawing cards without replacement makes successive draws: — Dependent. Without replacement, the sample space changes after each draw, making the events dependent.
  6. If A and B are independent, P(A|B) = ? — P(A). Independence means knowledge of B doesn't change P(A). So P(A|B) = P(A).
  7. From a class of 10 boys, 15 girls: P(boy | student selected) = ? — 2/5. P(boy) = 10/25 = 2/5. (No condition changes this since we are selecting from the whole class.)
  8. A permutation is: — An ordered arrangement. A permutation is an ordered arrangement. The order matters — ABC ≠ BAC.
  9. A combination is: — An unordered selection. A combination is an unordered selection. The order does not matter — ABC = BAC = CAB.
  10. 5! = ? — 120. 5! = 5 × 4 × 3 × 2 × 1 = 120.
  11. 0! = ? — 1. 0! = 1 by convention. This ensures combinatorial formulas work for edge cases.
  12. P(n, r) = ? — n!/(n−r)!. P(n, r) = n!/(n−r)!. This counts ordered arrangements of r items from n.
  13. C(n, r) = ? — n!/(r!(n−r)!). C(n, r) = n!/(r!(n−r)!). The r! in the denominator removes order.
  14. P(5, 3) = ? — 60. P(5, 3) = 5!/(5−3)! = 5!/2! = 120/2 = 60.
  15. C(5, 3) = ? — 10. C(5, 3) = 5!/(3!·2!) = 120/(6·2) = 10.
  16. C(10, 0) = ? — 1. C(n, 0) = n!/(0!·n!) = 1. There is exactly one way to choose nothing.
  17. C(7, 7) = ? — 1. C(n, n) = 1. There is exactly one way to choose all n items.
  18. How many subsets of size 2 does the set {1, 2, 3, 4, 5} have? — 10. C(5, 2) = 5!/(2!·3!) = 10. Each subset is an unordered pair.
  19. Expected value E(X) is: — The probability-weighted average of all outcomes. E(X) = Σ xᵢ·P(xᵢ). It is the long-run average outcome weighted by probability.
  20. A fair coin pays $2 for heads and −$1 for tails. E(X) = ? — $0.50. E(X) = 2·(1/2) + (−1)·(1/2) = 1 − 0.5 = $0.50.
  21. E(X) for a fair 6-sided die: — 3.5. E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5.
  22. A game is "fair" when E(X) = ? — 0. A fair game has E(X) = 0, meaning you expect to neither gain nor lose in the long run.
  23. Bayes' Theorem computes: — P(A|B) given P(B|A), P(A), and P(B). Bayes' Theorem: P(A|B) = P(B|A)·P(A)/P(B). It updates probability given new evidence.
  24. In E(X) = Σ xᵢ·P(xᵢ), what must the probabilities sum to? — 1. All probabilities in a probability distribution must sum to exactly 1.
  25. If you bet $10 on a coin flip (50/50), win $10 or lose $10, E(X) = ? — $0. E(X) = 10(0.5) + (−10)(0.5) = 5 − 5 = $0. This is a fair game.
  26. The formula P(A|B) = P(B|A)·P(A)/P(B) is known as: — Bayes' Theorem. This is Bayes' Theorem, named after Thomas Bayes. It expresses how to reverse conditional probabilities.