Experiments, outcomes, sample spaces and events; probabilities from equally likely outcomes

Set notation A′, A∩B, A∪B; complementary and mutually exclusive events

The addition law and the multiplication law

Conditional probability P(A|B) from formulae, Venn diagrams, two-way tables and tree diagrams

Testing whether two events are independent

Venn diagrams with two or three events, using counts or probabilities

Tree diagrams with and without replacement, and "reversed" conditional probabilities

1. Probability measures how likely an event is on a scale from 0 to 1, and almost every S1 probability question is solved by three laws (complement, addition, multiplication) kept honest by a Venn diagram or a tree diagram.

2. Equally likely outcomes: P(A)=n(A)/n(S). Complement: P(A′)=1-P(A); use it for "at least one".

3. Addition law: P(A∪B)=P(A)+P(B)-P(A∩B). If A and B are mutually exclusive, P(A∩B)=0.

4. Conditional probability: P(A|B)=P(A∩B)/P(B): the given event B becomes the new whole.

5. Independence: A and B are independent if and only if P(A∩B)=P(A)×P(B) (equivalently P(A|B)=P(A)). Test it with numbers and state a conclusion.

6. Trees: multiply along branches, add between paths. Without replacement, the second set of branches changes.

Probability questions in S1 are short to state and easy to get wrong. Almost every error comes from adding when you should multiply, dividing by the wrong total, or forgetting that a second pick changes when the first is not replaced. The cure is to be precise about what can happen.

An experiment is anything with an uncertain result: rolling a dice, choosing a person from a list, testing a component. An event such as "the score is even" is identified with the set of outcomes that make it happen. For a two-stage experiment, list the sample space as a grid, one stage across and one down: every cell is one equally likely outcome, so counting cells gives probabilities.

WORKED EXAMPLE · QUESTION

A fair four-sided spinner numbered 1, 2, 3, 4 is spun and a fair six-sided dice is rolled. The score is the product of the two numbers. G is the event "the score is at least 12" and D is the event "the score is odd". (a) Find P(G) and P(D). (b) Find P(G∩D). (c) Find P(G∪D).

SOLUTION

(a) Count the cells: P(G)=(8/24)= ⅓, P(D)=(6/24)= ¼.

(b) An odd product needs both numbers odd; only 3×5=15 is also at least 12: P(G∩D)=(1/24).

(c) Count cells in either event without counting 15 twice: (8/24)+(6/24)-(1/24)= (13/24). This is the addition law; counting the 13 marked cells directly agrees.

A and A′ are always mutually exclusive and exhaustive, which is why their probabilities add to 1. Three consequences are used constantly:

"At least one" is the complement of "none". For two fair dice, P(at least one six)=1-(5/6)²=(11/36). Adding ⅙+⅙ would count the double six twice.

A∩B′ means "A but not B", and P(A∩B′)=P(A)-P(A∩B).

"Neither A nor B" is A′∩B′, the complement of A∪B.

Any three of the four quantities give the fourth, and a Venn diagram shows all of them at once. To fill a two-event Venn diagram, start with the intersection, subtract it from P(A) and P(B) to get "A only" and "B only", then the outside is 1 minus the other three. Always check that the four regions add to 1. The number inside circle A but outside B is P(A∩B′), not P(A).

WORKED EXAMPLE · QUESTION

A customer at a corner shop is chosen at random. A is the event that the customer buys bread and B the event that the customer buys milk. P(A)=0.45, P(B)=0.3 and P(A∪B)=0.6. (a) Find P(A∩B). (b) Draw a Venn diagram showing the probabilities. (c) Find the probability that the customer buys exactly one of bread and milk. (d) Find P(A′∪B′).

SOLUTION

(a) Rearrange the addition law: P(A∩B)=0.45+0.3-0.6= 0.15.

(b) A only =0.45-0.15=0.3; B only =0.3-0.15=0.15; neither =1-0.6=0.4 (figure).

(c) Exactly one = A only + B only =0.3+0.15= 0.45.

(d) A′∪B′ is everything except A∩B: 1-0.15= 0.85.

Three overlapping circles make eight regions: the centre (all three), three "exactly two" regions, three "one only" regions and the outside. Fill from the centre outwards: all three first, then each "two only" region (pair total minus centre), then each "one only" region. The diagram can hold counts or probabilities; with counts, divide by the right total at the end, either the whole rectangle or, for a conditional probability, the given event. An unknown x is usually found from a total: the eight regions add to the number surveyed (or to 1).

WORKED EXAMPLE · QUESTION

Each of the 120 members of a sports club was asked whether they swim (S), cycle (C) or run (R). 6 do all three; 14 swim and cycle but do not run; 9 swim and run but do not cycle; x cycle and run but do not swim; 25 only swim; 18 only run; 12 do none of the three. The number who only cycle is twice the number who cycle and run but do not swim. (a) Find x. A member is chosen at random. Find the probability that the member (b) does exactly one of the activities, (c) swims, given that they cycle, (d) runs, given that they do at least two of the activities.

SOLUTION

(a) The eight regions add to 120: 6+14+9+x+25+2x+18+12=120⇒84+3x=120⇒ x=12.

(b) The "only" regions: (25+24+18)/120= (67/120).

(c) The given event C is the new total: n(C)=14+6+12+24=56, so P(S|C)=(14+6)/56= (5/14).

(d) At least two activities: 14+9+12+6=41; of these the runners are 9+12+6=27: (27/41).

Why this formula? Knowing that B happened shrinks the sample space to B. Of that, the part where A also happens is A∩B. So the denominator is always the probability (or count) of the given event. From a Venn diagram or a two-way table of counts, P(A|B)=n(A∩B)÷n(B). Note that P(A|B) and P(B|A) are usually different: same numerator, different denominators. The words "given that", "if it is known that" and "of those who" all signal a conditional probability.

WORKED EXAMPLE · QUESTION

A museum recorded whether each of its visitors on one day hired an audio guide.

One of these visitors is chosen at random. H is the event that the visitor hired a guide and D the event that the visitor is an adult. Find (a) P(H), (b) P(H|D), (c) P(D′|H). (d) Determine whether H and D are independent.

SOLUTION

(a) Column total over grand total: (90/250)= 0.36.

(b) Restrict to the adult row: (72/160)= 0.45.

(c) Restrict to the "hired" column: (18/90)= 0.2.

(d) P(H|D)=0.45≠0.36=P(H), so H and D are not independent. Equivalently P(H∩D)=72/250=0.288, but P(H)×P(D)=0.36×0.64=0.2304.

Show the numbers on both sides and finish with a sentence: "0.2=0.4×0.5, so A and B are independent".

If A and B are independent, so are A and B′, A′ and B, A′ and B′.

Mutually exclusive events with P(A),P(B)>0 are never independent: P(A∩B)=0 but P(A)P(B)>0. If one happens, the other cannot, which is the strongest possible dependence.

Questions sometimes tell you that events are independent so that you can multiply; use that to set up an equation.

WORKED EXAMPLE · QUESTION

Events A and B are such that P(A)=0.4, P(B)=p and P(A∪B)=0.7. (a) Find p if A and B are mutually exclusive. (b) Find p if instead A and B are independent. (c) In the case of part (b), find P(A|B′) and comment on your answer.

SOLUTION

(a) The overlap is 0: 0.7=0.4+p⇒ p=0.3.

(b) The overlap is 0.4p: 0.7=0.4+p-0.4p⇒0.6p=0.3⇒ p=0.5.

(c) P(A∩B′)=0.4-0.2=0.2 and P(B′)=0.5, so P(A|B′)=(0.2/0.5)= 0.4 = P(A). As expected: A and B are independent, so A and B′ are too, and knowing B did not happen does not change the chance of A.

A tree diagram organises an experiment in stages. Each set of branches from one point shows the outcomes of one stage, and the probabilities on those branches add to 1. Branches after the first stage carry conditional probabilities: the probability of the second outcome given the first. So:

multiply along a path to get the probability of that sequence (the multiplication law);

add the paths that make up an event (different paths are mutually exclusive);

for "at least one", use 1-P(none) rather than adding many paths.

With replacement (or independent stages) every set of branches is identical. Without replacement, the total goes down by 1 at each stage and so does the count of whatever was taken: the second stage depends on the first.

WORKED EXAMPLE · QUESTION

A jar contains 6 lemon sweets and 4 orange sweets. Kai takes a sweet at random and eats it, then takes a second sweet at random. Find the probability that (a) both sweets are the same flavour, (b) at least one sweet is orange, (c) the second sweet is lemon. (d) Given that the second sweet is lemon, find the probability that the first was orange.

SOLUTION

(a) (6/10)×(5/9)+(4/10)×(3/9)=(30/90)+(12/90)= (7/15).

(b) Complement of "both lemon": 1-(30/90)= ⅔.

(c) Two paths end in lemon: (30/90)+(24/90)=(54/90)= (3/5).

(d) OL path over all paths ending in L: (24/90)/(54/90)= (4/9).

The answer to (c) equals (6/10): with no information about the first pick, each pick is equally likely to be any of the 10 sweets. With replacement, (a) would be 0.6²+0.4²=0.52, more than (7/15)≈0.467: removing a sweet makes a second of the same flavour less likely.

A tree is drawn in time order: first stage, then second stage given the first. Questions often ask the reverse: given the second outcome, what is the probability of the first? Use P(first|second)=P(first∩second)/P(second): the numerator is one path, the denominator is the sum of every path that gives the second outcome. Mark the paths that satisfy the given event first; the answer is "the paths you want" ÷ "all marked paths".

WORKED EXAMPLE · QUESTION

A garden centre buys 60% of its seeds from supplier A and the rest from supplier B. A seed from A germinates with probability 0.9 and a seed from B with probability 0.75. A seed is chosen at random. (a) Find the probability that the seed germinates. (b) Given that the seed does not germinate, find the probability that it came from B. (c) Given that the seed germinates, find the probability that it came from A, to 3 significant figures.

SOLUTION

(a) P(G)=0.6×0.9+0.4×0.75=0.54+0.3= 0.84.

(b) P(B|G′)=(0.4×0.25)/(1-0.84)=(0.1/0.16)= 0.625.

(c) P(A|G)=(0.54/0.84)=(9/14)= 0.643 (3 s.f.).

WORKED EXAMPLE · QUESTION

A box contains 6 batteries, of which 2 are faulty. Batteries are taken at random, one at a time, and tested; a tested battery is not returned. Testing stops as soon as a faulty battery is found. T is the number of batteries tested. (a) Find P(T=1) and P(T=2). (b) Show that P(T=3)=⅕. (c) Find P(T>2). (d) Given that the first battery tested is not faulty, find the probability that exactly three batteries are tested. (e) Explain why T cannot be greater than 5.

SOLUTION

(a) P(T=1)=(2/6)= ⅓; P(T=2)=(4/6)×(2/5)= (4/15).

(b) Good, good, faulty: (4/6)×(3/5)×(2/4)=(24/120)=⅕ as required.

(c) 1-⅓-(4/15)= (2/5).

(d) The given event is T>1, probability 1-⅓=⅔, and T=3 lies inside it: (1/5)/(2/3)= (3/10).

(e) Only 4 batteries are good, so after at most 4 good ones the 5th tested must be faulty: T≤5. Check: P(T=4)=(2/15), P(T=5)=(1/15), and ⅓+(4/15)+⅕+(2/15)+(1/15)=1.

S1.5 Discrete random variables

Tree diagrams like Worked example 8 produce a probability distribution for a random variable such as T; its probabilities add to 1, and E(T) and Var(T) come next.

S2 The binomial distribution

A sequence of independent trials with the same probability of success, drawn as a tree with identical branches, is exactly the binomial model; the multiplication law gives pʳ(1-p)ⁿ⁻ʳ for each path.

S1.6 The Normal distribution

Every later probability, such as P(X>a) or a significance level, obeys the same complement and addition rules from this page.

1. P(A)=0.35, P(B)=0.5 and P(A∩B)=0.15. Find P(A∪B) and P(A|B).

2. Can two events, each with a non-zero probability, be both mutually exclusive and independent?

3. A bag holds 3 red and 2 white counters. Two are taken without replacement. Find the probability that they are different colours.

4. In a two-way table, 30 of the 80 people who own a bike also own a car. What is P(car | bike)?

5. P(A)=0.2, P(B|A)=0.6 and P(B|A′)=0.3. Find P(B) and P(A|B).

1. P(A∪B)=0.35+0.5-0.15=0.7. P(A|B)=0.15÷0.5=0.3.

2. No. Mutually exclusive gives P(A∩B)=0, but independence needs P(A∩B)=P(A)P(B), which is greater than 0.

3. RW or WR: (3/5)×(2/4)+(2/5)×¾=(6/20)+(6/20)=(3/5).

4. The given event is "owns a bike", so the denominator is 80: 30/80=0.375.

5. P(B)=0.2×0.6+0.8×0.3=0.12+0.24=0.36. P(A|B)=0.12÷0.36=⅓.