S1.3 Probability
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Statistics 1 (AS). Discrete random variables are on the S1.5 page; the binomial distribution is in S2. Permutations and combinations are not needed here.
What this page covers
7 points
- Experiments, outcomes, sample spaces and events; probabilities from equally likely outcomes
- Set notation \(A'\), \(A\cap B\), \(A\cup B\); complementary and mutually exclusive events
- The addition law and the multiplication law
- Conditional probability \(\text{P}(A\mid B)\) 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
In one minute
6 points
- 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.
- Equally likely outcomes: \(\text{P}(A)=\dfrac{n(A)}{n(S)}\). Complement: \(\text{P}(A')=1-\text{P}(A)\); use it for "at least one".
- Addition law: \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\). If \(A\) and \(B\) are mutually exclusive, \(\text{P}(A\cap B)=0\).
- Conditional probability: \(\text{P}(A\mid B)=\dfrac{\text{P}(A\cap B)}{\text{P}(B)}\): the given event \(B\) becomes the new whole.
- Independence: \(A\) and \(B\) are independent if and only if \(\text{P}(A\cap B)=\text{P}(A)\times\text{P}(B)\) (equivalently \(\text{P}(A\mid B)=\text{P}(A)\)). Test it with numbers and state a conclusion.
- Trees: multiply along branches, add between paths. Without replacement, the second set of branches changes.
Contents
14 sections
- What are sample spaces, events and equally likely outcomes?
- What do complementary and mutually exclusive mean?
- How does the addition law work?
- How do you use a Venn diagram with three events?
- What is conditional probability?
- How do you test whether two events are independent?
- How do tree diagrams work, with and without replacement?
- How do you reverse a conditional probability on a tree?
- Exam tips for S1.3
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What are sample spaces, events and equally likely outcomes?
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.
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 \(\text{P}(G)\) and \(\text{P}(D)\). (b) Find \(\text{P}(G\cap D)\). (c) Find \(\text{P}(G\cup D)\).
(a) Count the cells: \(\text{P}(G)=\dfrac{8}{24}=\) \(\tfrac13\), \(\text{P}(D)=\dfrac{6}{24}=\) \(\tfrac14\).
(b) An odd product needs both numbers odd; only \(3\times5=15\) is also at least 12: \(\text{P}(G\cap D)=\tfrac{1}{24}\).
(c) Count cells in either event without counting 15 twice: \(\dfrac{8}{24}+\dfrac{6}{24}-\dfrac{1}{24}=\) \(\tfrac{13}{24}\). This is the addition law; counting the 13 marked cells directly agrees.
What do complementary and mutually exclusive mean?
| Term | Meaning | Rule |
|---|---|---|
| Complement \(A'\) | "\(A\) does not happen": every outcome not in \(A\) | \(\text{P}(A')=1-\text{P}(A)\) |
| Intersection \(A\cap B\) | "\(A\) and \(B\)": outcomes in both | from a diagram, the addition law or the multiplication law |
| Union \(A\cup B\) | "\(A\) or \(B\) (or both)": outcomes in at least one | \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\) |
| Mutually exclusive | \(A\) and \(B\) cannot both happen: \(A\cap B=\varnothing\) | \(\text{P}(A\cap B)=0\), so \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)\) |
| Exhaustive | between them the events cover the whole sample space | \(\text{P}(A\cup B)=1\) |
\(A\) 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, \(\text{P}(\text{at least one six})=1-\left(\tfrac56\right)^2=\tfrac{11}{36}\). Adding \(\tfrac16+\tfrac16\) would count the double six twice.
- \(A\cap B'\) means "\(A\) but not \(B\)", and \(\text{P}(A\cap B')=\text{P}(A)-\text{P}(A\cap B)\).
- "Neither \(A\) nor \(B\)" is \(A'\cap B'\), the complement of \(A\cup B\).
Writing "mutually exclusive" when the question means "independent", or the reverse. Mutually exclusive is about the events overlapping (\(\text{P}(A\cap B)=0\)); independence is about one event changing the chance of the other. Two events with non-zero probabilities cannot be both.
How does the addition law work?
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 \(\text{P}(A)\) and \(\text{P}(B)\) 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 \(\text{P}(A\cap B')\), not \(\text{P}(A)\).
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. \(\text{P}(A)=0.45\), \(\text{P}(B)=0.3\) and \(\text{P}(A\cup B)=0.6\). (a) Find \(\text{P}(A\cap B)\). (b) Draw a Venn diagram showing the probabilities. (c) Find the probability that the customer buys exactly one of bread and milk. (d) Find \(\text{P}(A'\cup B')\).
(a) Rearrange the addition law: \(\text{P}(A\cap B)=0.45+0.3-0.6=\) 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) \(A'\cup B'\) is everything except \(A\cap B\): \(1-0.15=\) 0.85.
How do you use a Venn diagram with three events?
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).
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.
(a) The eight regions add to 120: \(6+14+9+x+25+2x+18+12=120\Rightarrow84+3x=120\Rightarrow\) \(x=12\).
(b) The "only" regions: \(\dfrac{25+24+18}{120}=\) \(\tfrac{67}{120}\).
(c) The given event \(C\) is the new total: \(n(C)=14+6+12+24=56\), so \(\text{P}(S\mid C)=\dfrac{14+6}{56}=\) \(\tfrac{5}{14}\).
(d) At least two activities: \(14+9+12+6=41\); of these the runners are \(9+12+6=27\): \(\tfrac{27}{41}\).
What is conditional probability?
Why this formula? Knowing that \(B\) happened shrinks the sample space to \(B\). Of that, the part where \(A\) also happens is \(A\cap B\). So the denominator is always the probability (or count) of the given event. From a Venn diagram or a two-way table of counts, \(\text{P}(A\mid B)=n(A\cap B)\div n(B)\). Note that \(\text{P}(A\mid B)\) and \(\text{P}(B\mid A)\) 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.
A museum recorded whether each of its visitors on one day hired an audio guide.
| Hired a guide | Did not hire | Total | |
|---|---|---|---|
| Adult | 72 | 88 | 160 |
| Child | 18 | 72 | 90 |
| Total | 90 | 160 | 250 |
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) \(\text{P}(H)\), (b) \(\text{P}(H\mid D)\), (c) \(\text{P}(D'\mid H)\). (d) Determine whether \(H\) and \(D\) are independent.
(a) Column total over grand total: \(\dfrac{90}{250}=\) 0.36.
(b) Restrict to the adult row: \(\dfrac{72}{160}=\) 0.45.
(c) Restrict to the "hired" column: \(\dfrac{18}{90}=\) 0.2.
(d) \(\text{P}(H\mid D)=0.45\ne0.36=\text{P}(H)\), so \(H\) and \(D\) are not independent. Equivalently \(\text{P}(H\cap D)=72/250=0.288\), but \(\text{P}(H)\times\text{P}(D)=0.36\times0.64=0.2304\).
How do you test whether two events are independent?
- Show the numbers on both sides and finish with a sentence: "\(0.2=0.4\times0.5\), 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 \(\text{P}(A),\text{P}(B)>0\) are never independent: \(\text{P}(A\cap B)=0\) but \(\text{P}(A)\text{P}(B)>0\). 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.
Events \(A\) and \(B\) are such that \(\text{P}(A)=0.4\), \(\text{P}(B)=p\) and \(\text{P}(A\cup B)=0.7\). (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 \(\text{P}(A\mid B')\) and comment on your answer.
(a) The overlap is 0: \(0.7=0.4+p\Rightarrow\) \(p=0.3\).
(b) The overlap is \(0.4p\): \(0.7=0.4+p-0.4p\Rightarrow0.6p=0.3\Rightarrow\) \(p=0.5\).
(c) \(\text{P}(A\cap B')=0.4-0.2=0.2\) and \(\text{P}(B')=0.5\), so \(\text{P}(A\mid B')=\dfrac{0.2}{0.5}=\) 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 \(A\).
How do tree diagrams work, with and without replacement?
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 \(1-\text{P}(\text{none})\) 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.
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.
(a) \(\tfrac{6}{10}\times\tfrac59+\tfrac{4}{10}\times\tfrac39=\tfrac{30}{90}+\tfrac{12}{90}=\) \(\tfrac{7}{15}\).
(b) Complement of "both lemon": \(1-\tfrac{30}{90}=\) \(\tfrac23\).
(c) Two paths end in lemon: \(\tfrac{30}{90}+\tfrac{24}{90}=\tfrac{54}{90}=\) \(\tfrac35\).
(d) OL path over all paths ending in L: \(\dfrac{24/90}{54/90}=\) \(\tfrac49\).
The answer to (c) equals \(\tfrac{6}{10}\): with no information about the first pick, each pick is equally likely to be any of the 10 sweets. With replacement, (a) would be \(0.6^2+0.4^2=0.52\), more than \(\tfrac{7}{15}\approx0.467\): removing a sweet makes a second of the same flavour less likely.
Keeping the denominator at 10 on the second branches, or reducing the denominator but forgetting to reduce the count of the flavour already taken. After a lemon is removed, it is 5 lemons out of 9, not 6 out of 9 or 5 out of 10.
How do you reverse a conditional probability on a tree?
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 \(\text{P}(\text{first}\mid\text{second})=\dfrac{\text{P}(\text{first}\cap\text{second})}{\text{P}(\text{second})}\): 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".
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.
(a) \(\text{P}(G)=0.6\times0.9+0.4\times0.75=0.54+0.3=\) 0.84.
(b) \(\text{P}(B\mid G')=\dfrac{0.4\times0.25}{1-0.84}=\dfrac{0.1}{0.16}=\) 0.625.
(c) \(\text{P}(A\mid G)=\dfrac{0.54}{0.84}=\dfrac{9}{14}=\) 0.643 (3 s.f.).
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 \(\text{P}(T=1)\) and \(\text{P}(T=2)\). (b) Show that \(\text{P}(T=3)=\tfrac15\). (c) Find \(\text{P}(T>2)\). (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.
(a) \(\text{P}(T=1)=\tfrac26=\) \(\tfrac13\); \(\text{P}(T=2)=\tfrac46\times\tfrac25=\) \(\tfrac{4}{15}\).
(b) Good, good, faulty: \(\tfrac46\times\tfrac35\times\tfrac24=\tfrac{24}{120}=\tfrac15\) as required.
(c) \(1-\tfrac13-\tfrac{4}{15}=\) \(\tfrac25\).
(d) The given event is \(T>1\), probability \(1-\tfrac13=\tfrac23\), and \(T=3\) lies inside it: \(\dfrac{1/5}{2/3}=\) \(\tfrac{3}{10}\).
(e) Only 4 batteries are good, so after at most 4 good ones the 5th tested must be faulty: \(T\le5\). Check: \(\text{P}(T=4)=\tfrac{2}{15}\), \(\text{P}(T=5)=\tfrac{1}{15}\), and \(\tfrac13+\tfrac{4}{15}+\tfrac15+\tfrac{2}{15}+\tfrac{1}{15}=1\).
In part (d), do not draw a new tree. The given information picks out part of the original sample space; divide the probability of the event you want (inside that part) by the probability of the whole part.
Exam tips for S1.3
Whenever there are two or three events, draw a Venn diagram; whenever there are stages, draw a tree. Write every region or branch value on it, and check that the regions (or the branches from each point) add to 1.
Find the words "given that" (or "if", "of those who") and underline the event that follows: its probability or count is the denominator. Write \(\text{P}(A\mid B)=\text{P}(A\cap B)/\text{P}(B)\) before substituting.
Calculate both \(\text{P}(A\cap B)\) and \(\text{P}(A)\times\text{P}(B)\), write them side by side with = or ≠, and state the conclusion in words. "They look unrelated" is not a test.
If the item is not put back, change both the numerator and the denominator on later branches. Words such as "eats", "keeps", "chooses two different" and "is not returned" all mean without replacement.
Give probabilities as exact fractions in their simplest form or as exact decimals. If you round, give 3 significant figures and keep full accuracy until the final line.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Mutually exclusive and independent mean the same thing. | Mutually exclusive events cannot happen together; independent events happen together with probability \(\text{P}(A)\text{P}(B)>0\). | With non-zero probabilities, mutually exclusive events are always dependent. |
| \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)\) always. | The overlap is counted twice. | Subtract \(\text{P}(A\cap B)\); it is only zero for mutually exclusive events. |
| \(\text{P}(A\mid B)=\text{P}(B\mid A)\). | Same numerator, different denominators. | In Worked example 7, \(\text{P}(G'\mid B)=0.25\) but \(\text{P}(B\mid G')=0.625\). |
| \(\text{P}(A\mid B)=\text{P}(A\cap B)\div\text{P}(A)\). | That divides by the wrong event. | Divide by the probability of the given event, \(\text{P}(B)\). |
| The number inside circle \(A\) on a Venn diagram is \(\text{P}(A)\). | It is only the part of \(A\) outside the other circles. | \(\text{P}(A)\) is the total of every region inside circle \(A\). |
| P(at least one six from two dice) \(=\tfrac16+\tfrac16\). | The double six is counted twice. | \(1-\left(\tfrac56\right)^2=\tfrac{11}{36}\). |
How this topic connects
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 \(\text{E}(T)\) and \(\text{Var}(T)\) 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 \(p^r(1-p)^{n-r}\) for each path.
S1.6 The Normal distribution
Every later probability, such as \(\text{P}(X>a)\) or a significance level, obeys the same complement and addition rules from this page.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\text{P}(A)=n(A)/n(S)\) | probability from equally likely outcomes | No – learn it |
| \(\text{P}(A')=1-\text{P}(A)\) | complement; "at least one" = 1 − P(none) | No – learn it |
| \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\) | addition law | Yes |
| \(\text{P}(A\cap B)=\text{P}(A\mid B)\text{P}(B)\) | multiplication law; multiplying along tree branches | Yes |
| \(\text{P}(A\mid B)=\text{P}(A\cap B)/\text{P}(B)\) | conditional probability | Yes |
| \(\text{P}(A\cap B)=0\) | condition for mutually exclusive events | No – learn it |
| \(\text{P}(A\cap B)=\text{P}(A)\times\text{P}(B)\iff\text{P}(A\mid B)=\text{P}(A)\) | condition (and test) for independent events | No – learn it |
| \(\text{P}(A\cap B')=\text{P}(A)-\text{P}(A\cap B)\) | "\(A\) but not \(B\)", the \(A\)-only region of a Venn diagram | No – derive it |
Check your understanding
1. \(\text{P}(A)=0.35\), \(\text{P}(B)=0.5\) and \(\text{P}(A\cap B)=0.15\). Find \(\text{P}(A\cup B)\) and \(\text{P}(A\mid B)\).
\(\text{P}(A\cup B)=0.35+0.5-0.15=0.7\). \(\text{P}(A\mid B)=0.15\div0.5=0.3\).
2. Can two events, each with a non-zero probability, be both mutually exclusive and independent?
No. Mutually exclusive gives \(\text{P}(A\cap B)=0\), but independence needs \(\text{P}(A\cap B)=\text{P}(A)\text{P}(B)\), which is greater than 0.
3. A bag holds 3 red and 2 white counters. Two are taken without replacement. Find the probability that they are different colours.
RW or WR: \(\tfrac35\times\tfrac24+\tfrac25\times\tfrac34=\tfrac{6}{20}+\tfrac{6}{20}=\tfrac35\).
4. In a two-way table, 30 of the 80 people who own a bike also own a car. What is P(car | bike)?
The given event is "owns a bike", so the denominator is 80: \(30/80=0.375\).
5. \(\text{P}(A)=0.2\), \(\text{P}(B\mid A)=0.6\) and \(\text{P}(B\mid A')=0.3\). Find \(\text{P}(B)\) and \(\text{P}(A\mid B)\).
\(\text{P}(B)=0.2\times0.6+0.8\times0.3=0.12+0.24=0.36\). \(\text{P}(A\mid B)=0.12\div0.36=\tfrac13\).
Specification checklist
- (S1 3.1) I can describe the sample space of an experiment, list or tabulate its outcomes, and find probabilities by counting equally likely outcomes.
- (S1 3.1) I can use set notation for complements, unions and intersections, and use \(\text{P}(A')=1-\text{P}(A)\), including for "at least one" problems.
- (S1 3.1) I can recognise mutually exclusive events and use the fact that the probability of their intersection is zero.
- (S1 3.2) I can use the addition law \(\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\), working in either direction.
- (S1 3.2) I can use the multiplication law and find conditional probabilities \(\text{P}(A\mid B)\) from formulae, two-way tables and diagrams.
- (S1 3.3) I can test whether two events are independent, showing the numerical comparison and stating a conclusion, and use independence to find unknown probabilities.
- (S1 3.4) I can draw and use Venn diagrams for two or three events, with counts or probabilities, including ones with an unknown to find.
- (S1 3.4) I can draw and use tree diagrams for selections with and without replacement, and find conditional probabilities from Venn and tree diagrams, including reversed conditionals.