Flashcards · 0580 · Topic 8

Probability flashcards

24 cards on Topic 8, subtopics 8.1 to 8.4. Pick a subtopic or study them all.

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The probability scale: From 0 (impossible) to 1 (certain). 0.5 is an even chance. Give probabilities as fractions, decimals or percentages, never more than 1.
Probability of an event when all outcomes are equally likely: P = number of favourable outcomes ÷ total number of possible outcomes
Probability of an event not happening: 1 − the probability that it happens. The probabilities of all the possible outcomes sum to 1.
A spinner lands on red 0.35, blue 0.2, green x and yellow 2x. Find x.: 0.35 + 0.2 + 3x = 1, so 3x = 0.45 and x = 0.15.
Venn diagram: S only 9, both S and T 5, T only 11, neither 7. One person is chosen at random. Find the probability they are in S.: Total 9 + 5 + 11 + 7 = 32. In S: 9 + 5 = 14. 14/32 = 7/16
Probability notation P(A) and P(A′)Extended: P(A): the probability of event A. P(A′): the probability of not A, so P(A′) = 1 − P(A).

Relative frequency: number of times the outcome happened ÷ total number of trials An experimental estimate of the probability.
Which relative frequency gives the best estimate of a probability?: The one from the most trials. Relative frequency tends to settle as the number of trials grows.
Expected frequency: expected frequency = probability × number of trials It is what you expect on average; the real result is usually close, not exact.
A fair ten-sided dice (1 to 10) is rolled 150 times. How many times do you expect a multiple of 3?: Multiples of 3: 3, 6, 9, so P = 3/10. 3/10 × 150 = 45
Fair, biased and random: Fair: every outcome is equally likely. Biased: the outcomes are not equally likely. Random: chosen by chance, with no outcome favoured.
After many trials, how do you decide whether a spinner is biased?: Compare the relative frequencies from a large number of trials with the fair probabilities. Big differences suggest bias; say "suggests", not "proves".

When do you multiply probabilities and when do you add them?: Multiply for A AND B when the events are independent. Add for A OR B when the events are mutually exclusive.
Tree diagram rules: Branches from the same point sum to 1. Multiply along a path for its probability. Add the probabilities of the different paths that give the outcome you want.
Two fair dice are rolled. Find the probability that the scores add to 7.: Sample space: 36 equally likely outcomes. Six give 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6/36 = 1/6
A bag has 5 red and 3 blue counters. One is taken, replaced, then another taken. Find P(both red).: 5/8 × 5/8 = 25/64
A fair dice is rolled twice. Find the probability of at least one six.: 1 − P(no six) = 1 − (5/6)² = 1 − 25/36 = 11/36
A bag has 5 red and 3 blue counters. Two are taken WITHOUT replacement. Find P(both red).Extended: 5/8 × 4/7 = 20/56 = 5/14 After one red is removed, 4 of the 7 left are red.
Venn diagram notation P(A ∩ B) and P(A ∪ B)Extended: P(A ∩ B): the probability of A and B. P(A ∪ B): the probability of A or B (or both).

Conditional probability: the methodExtended: Restrict attention to the group already known to have happened, then find the fraction of that group with the outcome you want. No formula or P(B|A) notation is required.
12 of 30 students play chess, and 5 of those also play tennis. A chess player is chosen at random. Find P(plays tennis).Extended: 5/12. Only the 12 chess players can be chosen.
Two-way table: 16 girls, 11 of whom take the bus (30 students in all). A girl is chosen at random. Find P(she takes the bus).Extended: 11/16. Divide by the number of girls, not by 30.
On a tree diagram for picking without replacement, what are the second-stage branches?Extended: Conditional probabilities: each is the probability of the second outcome given the first. After a red from 5 red and 3 blue, P(red) = 4/7.
Exam trap in conditional probabilityExtended: Dividing by the whole total instead of by the size of the group already known to have happened.