Skip to content
Marlbridge

Flashcards · 4040 · Topic 11

Elementary ideas of probability flashcards

12 cards on Topic 11. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 12 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

Topic 11: Elementary ideas of probability

What are the sample space and an event?
The sample space S is the set of all possible outcomes of an experiment. An event is the set of outcomes you are interested in.
How do you find P(A) when outcomes are equally likely?
P(A) = n(A) ÷ n(S), and 0 ≤ P(A) ≤ 1. Example: P(more than 4 on a fair die) = 2/6 = 1/3.
An event happens 36 times in 150 trials. Estimate its probability and the number expected in 400 trials
Relative frequency = 36 ÷ 150 = 0.24. Expected number = probability × trials = 0.24 × 400 = 96.
What is the complement of an event?
A′, the event that A does not happen. P(A′) = 1 − P(A).
What are mutually exclusive events, and what rule do they follow?
Events that cannot both happen in one trial, so A ∩ B = Ø and P(A ∪ B) = P(A) + P(B). A full set of exclusive outcomes has probabilities adding to 1.
P(A) = 0.6, P(B) = 0.35 and P(neither) = 0.1. Find P(A ∩ B)
P(A ∪ B) = 1 − 0.1 = 0.9. Then P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.35 − 0.9 = 0.05.
Could events with P(A) = 0.6 and P(B) = 0.5 be mutually exclusive?
No. If they were, P(A ∪ B) = 0.6 + 0.5 = 1.1, which is more than 1.
How do you test two events for independence?
Check whether P(A ∩ B) = P(A) × P(B). Example: 50 members, 20 in choir, 10 in band, 4 in both: 20/50 × 10/50 = 0.08 = 4/50, so they are independent.
A bag has 4 black and 5 white counters. Find P(both black) with and without replacement
With: 4/9 × 4/9 = 16/81. Without: 4/9 × 3/8 = 12/72 = 1/6. Without replacement the total and the count of the colour taken both fall by one.
Each component fails with probability 0.3, independently. Find P(at least one of three fails)
P(all work) = 0.7³ = 0.343, so P(at least one fails) = 1 − 0.343 = 0.657.
What are the rules for tree diagrams?
Branches from one point add to 1. Multiply along a path; add the paths that make up the event, since different paths are exclusive. All path probabilities add to 1.
How do you find the probability of dependent picks in order?
P(X then Y) = P(X) × P(Y given X). Example: 2 red and 6 yellow, no replacement, first red on the second pick: 6/8 × 2/7 = 3/14.