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Flashcards · 0479 · Topic 7

Probability flashcards

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Topic 7: Probability

What do P(A′), P(A ∩ B) and P(A ∪ B) mean?
P(A′): A does not happen. P(A ∩ B): both A and B happen. P(A ∪ B): A or B or both happen.
How do you find a probability when outcomes are equally likely?
P = number of favourable outcomes ÷ total number of outcomes. Every probability lies between 0 (impossible) and 1 (certain), as a fraction, decimal or percentage.
What is the addition rule for any two events?
P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Example: P(A) = 2/5, P(B) = 1/3, P(A ∩ B) = 1/10 → 12/30 + 10/30 − 3/30 = 19/30.
How do with-replacement and without-replacement picks differ?
With replacement the probabilities are the same at every pick. Without replacement the total and the count of the type taken both fall by one. Example, 3 grey and 5 black socks, both grey: with 3/8 × 3/8 = 9/64; without 3/8 × 2/7 = 3/28.
What is the quickest way to find P(at least one)?
P(at least one) = 1 − P(none). Example: three independent attempts, each succeeding with probability 0.2 → 1 − 0.8³ = 1 − 0.512 = 0.488.
What are mutually exclusive events?
Events that cannot happen together, so P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B).
What are independent events, and how do you test for independence?
One happening does not change the probability of the other, so P(A ∩ B) = P(A) × P(B). Test: if P(A) × P(B) equals P(A ∩ B), they are independent. State the conclusion in context.
Can two mutually exclusive events with non-zero probabilities be independent?
No. For exclusive events P(A ∩ B) = 0, but P(A) × P(B) is not 0.
How do you find a conditional probability without a formula?
Keep only the group named after "given"; that is the denominator. Count how many of that group also meet the event. Example: 27 chess players, 12 of them adults → P(adult given chess player) = 12/27 = 4/9.
What are the rules for using a tree diagram?
Branches from one point add to 1. Multiply along a path for that outcome; add the paths that make up the event. All end outcomes add to 1.
Which probability methods does 0479 not require?
Permutation and combination formulas (count by listing, or multiply the choices at each pick), and the formula P(A | B) = P(A ∩ B)/P(B) (use the restricted-group method instead).

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