Flashcards · 9660 · Topic 3

Unit PSM1 – S1: Statistics (International AS) flashcards

17 cards on Topic 3, subtopics S1.1 to S1.3. Pick a subtopic or study them all.

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Addition law for two events, and for mutually exclusive events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
Multiplication law and conditional probability: P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B). P(B | A) = P(A ∩ B)/P(A).
P(A) = 0.45, P(B) = 0.35 and P(A ∩ B) = 0.2. Are A and B independent?: P(A) × P(B) = 0.45 × 0.35 = 0.1575, but P(A ∩ B) = 0.2. These differ, so A and B are not independent.
Can two events with non-zero probabilities be both mutually exclusive and independent?: No. If they are mutually exclusive, one happening rules out the other (P(A ∩ B) = 0), but independence needs P(A ∩ B) = P(A)P(B) > 0.
A bag has 5 red and 3 blue counters. Two are taken without replacement. Find P(both red).: 5/8 × 4/7 = 20/56 = 5/14. Without replacement, reduce the numerator and denominator after each pick.
Three independent switches work with probabilities 0.9, 0.8 and 0.7. Find P(at least one fails).: P(all work) = 0.9 × 0.8 × 0.7 = 0.504. P(at least one fails) = 1 − 0.504 = 0.496.

Valid discrete distribution, and the mean E(X): 0 ≤ P(X = x) ≤ 1 and Σ P(X = x) = 1. E(X) = μ = Σ x p. For a function, E(g(X)) = Σ g(x) p (never put E(X) into g).
Variance of a discrete random variable: Var(X) = E((X − μ)²) = E(X²) − μ², where E(X²) = Σ x² p. Standard deviation σ = √Var(X).
E(aX + b) and Var(aX + b): E(aX + b) = aE(X) + b. Var(aX + b) = a²Var(X). So Var(−X) = Var(X).
E(aX ± bY) and Var(aX ± bY) for independent X and Y: E(aX ± bY) = aE(X) ± bE(Y). Var(aX ± bY) = a²Var(X) + b²Var(Y). Variances always add; the variance formula needs independence.
W takes 10, 20, 30 with probabilities 0.2, 0.5, 0.3. Find E(W) and Var(W).: E(W) = 2 + 10 + 9 = 21. E(W²) = 20 + 200 + 270 = 490. Var(W) = 490 − 441 = 49, so σ = 7.
Total of n independent copies of X compared with nX: means and variances: X₁ + … + Xₙ: mean nE(X), variance nVar(X). nX (one value scaled): mean nE(X), variance n²Var(X).

Bernoulli trial and Bernoulli distribution: A single trial with exactly two outcomes: success (probability p) and failure (1 − p). Mean p, variance p(1 − p).
Binomial distribution B(n, p) and why its mean and variance are np and np(1 − p): X ~ B(n, p) is the number of successes in n independent, identical Bernoulli trials. It is a sum of n Bernoulli variables, so its mean is n × p and its variance is n × p(1 − p).
Four conditions for a binomial model: 1. A fixed number of trials n. 2. Two possible outcomes per trial. 3. A constant probability of success p. 4. Independent trials. Sampling without replacement from a small population breaks independence.
Components are defective with probability 0.1. In a sample of 8, find P(exactly 2 defective).: P(X = 2) = ⁸C₂ (0.1)²(0.9)⁶ = 28 × 0.01 × 0.531441 ≈ 0.149.
Using cumulative binomial tables for P(X ≥ r): Tables give P(X ≤ x). Use P(X ≥ r) = 1 − P(X ≤ r − 1). Example: X ~ B(10, 0.3) has P(X ≤ 3) = 0.6496, so P(X ≥ 4) = 0.3504.