Probability

Syllabus coverage

Evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations
Use addition and multiplication of probabilities, as appropriate, in simple cases
Understand the meaning of exclusive and independent events, including determination of whether events A and B are independent by comparing the values of $P (A \cap B)$ and $P (A) \times P (B)$
Calculate and use conditional probabilities in simple cases, including situations that can be represented by a sample space of equiprobable elementary events or by a tree diagram, and the use of $P (A \cap B) = P (A) P (B ∣ A)$

Part of the AS & A Level Maths syllabus 9709.

Three views of one idea: a two-way table, a Venn diagram and a tree diagram, each showing how a conditional probability restricts the group you divide by.

Events, complements and the sample space

A probability question is a question about a sample space: the list of everything that could happen. One item on the list is an outcome, a collection of outcomes is an event, and $P (A)$ is the probability that event $A$ happens. The sample space is written $ξ$ , and the probability of the whole of it is 1.

The complement of $A$ , written $A^{'}$ , is everything in the sample space that is not in $A$ . Between them they cover every outcome, so $P (A) + P (A^{'}) = 1$ and therefore $P (A^{'}) = 1 - P (A)$ . Every 'at least one' question is really a complement question: counting the outcomes where the event does not happen is quicker than counting the ones where it does.

Two events combine in two ways. $A \cap B$ is the intersection, read as 'A and B', and it holds the outcomes that are in both. $A \cup B$ is the union, read as 'A or B', and it holds the outcomes in at least one of them - the word 'or' in probability always includes the case where both happen.

Notation	Read as	What it stands for
$P (A)$	the probability of A	the outcomes in A as a fraction of the sample space
$A^{'}$	the complement of A	every outcome that is not in A, with $P (A^{'}) = 1 - P (A)$
$A \cap B$	A and B	the outcomes that are in both A and B
$A \cup B$	A or B	the outcomes in A, in B, or in both
$P (A ∣ B)$	A given B	the probability of A when the sample space is restricted to B

'Or' includes 'both'. A question that wants one or the other but not both says so - 'exactly one of them' - and then you subtract the intersection.

Equiprobable elementary events: counting is the whole method

When every outcome in the sample space is equally likely, a probability is nothing more than a count: $P (A) = \frac{the number of outcomes in A}{the number of outcomes in ξ}$ . A fair die has six equally likely outcomes, a fair coin two, and a well-shuffled pack of cards fifty-two. Almost everything in this topic rests on that sentence, so the first move in any question is to decide what the equally likely outcomes are and how many of them there are.

Two fair dice give 36 equally likely outcomes, not 11. Writing the totals 2 to 12 down and treating them as equally likely is the single commonest mistake in the topic: a total of 7 can happen in six ways and a total of 2 in only one, so the totals are not equally likely even though each die is fair. Count pairs of faces, not totals, and the difficulty disappears.

Once the counts are written down the arithmetic is a fraction. $P (total = 7) = \frac{6}{36} = \frac{1}{6}$ , and $P (total \geq 10) = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6}$ . Write the fraction before you simplify it: the working is what the mark scheme follows, and a correct count that simplifies wrongly still scores the method mark.

Total score	Number of ways	Probability
2	1	$\frac{1}{36}$
3	2	$\frac{2}{36} = \frac{1}{18}$
4	3	$\frac{3}{36} = \frac{1}{12}$
5	4	$\frac{4}{36} = \frac{1}{9}$
6	5	$\frac{5}{36}$
7	6	$\frac{6}{36} = \frac{1}{6}$
8	5	$\frac{5}{36}$
9	4	$\frac{4}{36} = \frac{1}{9}$
10	3	$\frac{3}{36} = \frac{1}{12}$
11	2	$\frac{2}{36} = \frac{1}{18}$
12	1	$\frac{1}{36}$

A probability is a fraction of a sample space, so it can never be negative and never greater than 1. If an answer comes out as 1.2, the counting has gone wrong - look for outcomes counted twice.

The probability scale on the left, and on the right the 36 equally likely outcomes of throwing two fair dice. Six of the 36 cells total 7.

Counting outcomes with permutations and combinations

When the sample space is too big to list, count it. The multiplication principle does most of the work: if a first choice can be made in $m$ ways and a second in $n$ ways, the two together can be made in $m \times n$ ways. Three counters drawn one at a time from a bag of 10 give $10 \times 9 \times 8$ ordered selections, and two dice give $6 \times 6 = 36$ outcomes - the same 36 you would have written out by hand.

The number of ways of arranging $n$ distinct objects in a line is $n!$ , because there are $n$ choices for the first place, $n - 1$ for the second, and so on down to 1. The number of ways of choosing $r$ objects from $n$ when the order does not matter is $(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ , and your calculator's nCr button computes it. For example $(\binom{9}{3}) = \frac{9!}{3! 6!} = 84$ committees of 3 can be chosen from 9 people, and $(\binom{4}{2}) = 6$ pairs can be chosen from 4 women.

The one rule that must not be broken is that the top and the bottom of the fraction are counted the same way. If the sample space is a number of combinations - all the hands of 3 that could be chosen - then the event must be counted in combinations too. Mixing $(\binom{9}{3})$ underneath with $9 \times 8$ on top counts the same committee several times over, so the fraction that comes out is not the probability it looks like.

Order matters (arrangements, finishing orders, passwords): count with $n!$ or with the multiplication principle
Order does not matter (committees, hands of cards, selections from a bag): count with $(\binom{n}{r})$
The sample space of hands of 5 cards from 52 is $(\binom{52}{5}) = 2 598 960$
The sample space of ordered selections of 3 counters from 10 is $10 \times 9 \times 8 = 720$ ; unordered it is $(\binom{10}{3}) = 120$
A fraction whose top and bottom are counted in different ways is wrong however small the numbers are

Every combination is equally likely when the objects are drawn at random, which is exactly why counting works. 'At random' is the phrase in the question that licenses the division.

Mutually exclusive events and the addition rule

Two events are mutually exclusive (or disjoint) when they cannot both happen. There is no outcome in both, so $A \cap B = \emptyset$ and $P (A \cap B) = 0$ . Throwing a head and throwing a tail with one coin are mutually exclusive; drawing a red counter and drawing a blue counter from a bag that holds only red and blue counters are mutually exclusive; scoring a total of 7 and scoring a total of 9 with two dice are mutually exclusive.

For mutually exclusive events the probabilities add: $P (A \cup B) = P (A) + P (B)$ , and the same rule extends to any number of them. When the events are also exhaustive - between them they cover every outcome in the sample space - the probabilities add to exactly 1, which is a useful way to find a missing probability.

When events are not mutually exclusive you must not simply add, because the outcomes in both are then counted twice. The general formula $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ takes the overlap out again. Explicit use of that formula is not required at 9709, but it is the safest check on a Venn diagram answer, and it explains why the addition rule needs the word 'exclusive' in front of it.

The situation	The rule	An example
Mutually exclusive	$P (A \cup B) = P (A) + P (B)$	a fair die shows 1 or 2: $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$
Mutually exclusive and exhaustive	the probabilities add to 1	a fair die shows an odd number or an even number
Not mutually exclusive	subtract the overlap before adding	with $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.2$ , $P (A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$

Mutually exclusive is a statement about the events: they cannot both happen. Independence is a statement about the numbers:

P (A \cap B) = P (A) \times P (B)

. An exclusive pair with both probabilities greater than zero is never independent.

A two-set Venn diagram. Mutually exclusive events have nothing in the lens between them, so the two circles do not overlap.

Independent events and the multiplication rule

Two events are independent when knowing that one of them has happened tells you nothing at all about the other. Tosses of a fair coin are independent, the faces of two separate dice are independent, and so are draws from a bag when each counter is put back before the next one is drawn.

For independent events the probabilities multiply: $P (A \cap B) = P (A) \times P (B)$ , and for three independent events the rule extends to $P (A \cap B \cap C) = P (A) P (B) P (C)$ . So the probability of three heads in three tosses is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ .

The distinction that decides most questions is with or without replacement. With replacement the bag is restored after each draw, so the probabilities never change and the draws are independent. Without replacement the first draw changes the contents of the bag, so the second probability is a conditional probability - and the multiplication rule becomes $P (A \cap B) = P (A) \times P (B ∣ A)$ .

Take a bag holding 3 red and 2 blue counters and draw two. With replacement, $P (both red) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ . Without replacement one red counter has gone, so only 2 of the remaining 4 are red and $P (both red) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ . Both answers are smaller than the probability of the first counter being red, but they are different numbers, and that difference is the whole point of the topic.

Multiplying is only justified when the events are independent, or when the second factor is written as a conditional probability.

\frac{5}{8} \times \frac{4}{7}

is correct for two draws without replacement;

\frac{5}{8} \times \frac{5}{8}

is not.

The same bag of 3 red and 2 blue counters drawn with replacement and without. The second-stage fractions stay at fifths in the first tree and change to quarters in the second.

Deciding whether two events are independent

Independence is decided by comparing two numbers: $P (A \cap B)$ and $P (A) \times P (B)$ . If the two are equal the events are independent; if they differ, even slightly, the events are dependent. Nothing else decides it - not the size of the probabilities, not the wording of the question, and not how plausible independence feels.

Suppose $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.2$ . Then $P (A) \times P (B) = 0.5 \times 0.4 = 0.2$ , which equals $P (A \cap B)$ , so A and B are independent. Now change $P (A \cap B)$ to 0.3 and the same two events are dependent, because $0.5 \times 0.4 = 0.2 \neq 0.3$ .

The same test works on a table of counts. In the group of 60 students used later on this page, $P (boy) = \frac{30}{60} = 0.5$ and $P (walks) = \frac{24}{60} = 0.4$ , so $P (boy) \times P (walks) = 0.2$ . But 18 students are boys who walk, so $P (boy \cap walks) = \frac{18}{60} = 0.3$ . The numbers are different, so being a boy and walking to school are not independent: boys are more likely than girls to walk.

There is a second, equivalent test which is often quicker in words: A and B are independent exactly when $P (A ∣ B) = P (A)$ , that is when restricting attention to B leaves the probability of A unchanged. For the 60 students, $P (boy ∣ walks) = \frac{18}{24} = 0.75$ , and $P (boy) = 0.5$ . Those differ, which confirms what the first test said.

The test	Independent when	Dependent when
$P (A \cap B)$ against $P (A) \times P (B)$	the two values are equal	the two values are not equal
$P (A ∣ B)$ against $P (A)$	$P (A ∣ B) = P (A)$	$P (A ∣ B) \neq P (A)$
A mutually exclusive pair	never, unless one probability is 0	always, when both probabilities are greater than 0

Do not judge independence from how big the probabilities are.

P (A) = 0.9

and

P (B) = 0.9

with

P (A \cap B) = 0.81

are independent, while

P (A) = 0.1

and

P (B) = 0.2

with

P (A \cap B) = 0.1

are not.

Conditional probability: restricting the sample space

$P (A ∣ B)$ is the probability that A happens given that B has happened. The phrase 'given that' is the whole of the definition: the sample space shrinks to B, and you count only the part of A that lies inside B. That gives $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ , provided $P (B) > 0$ .

A table of counts makes this concrete. In a group of 60 students, 24 walk to school and the other 36 travel by bus. There are 30 boys in the group, and 18 of the boys walk to school, so 6 girls walk. Now ask for the probability that a student who walks to school is a boy. The condition 'walks to school' fixes the group at 24 students, and 18 of those are boys, so $P (boy ∣ walks) = \frac{18}{24} = \frac{3}{4}$ . The denominator is 24, not 60. Compare $P (boy) = \frac{30}{60} = \frac{1}{2}$ for a student chosen at random from the whole group: the condition has changed the probability, so the two events are not independent.

The formula and the counting must agree, and that is the check to use when a question gives probabilities instead of counts. Here $P (boy \cap walks) = \frac{18}{60} = 0.3$ and $P (walks) = \frac{24}{60} = 0.4$ , so $P (boy ∣ walks) = \frac{0.3}{0.4} = 0.75$ , the same $\frac{3}{4}$ . When only probabilities are given, that division is the only route.

The event after the bar is the group you are told about, and it goes on the bottom of the fraction. Reversing the bar changes the answer:

P (boy ∣ walks) = \frac{18}{24}

, but

P (walks ∣ boy) = \frac{18}{30}

. Both use the same 18 and different denominators.

Tree diagrams and the reverse conditional probability

A tree diagram handles dependent events cleanly. Each branch carries the probability of that outcome at that stage, the probability of a route through the tree is the product of the branches along it, and the branches leaving any point add to 1. The product along a route is the multiplication rule in its general form, $P (A \cap B) = P (A) \times P (B ∣ A)$ , which is what makes trees correct without replacement as well as with it.

Take a bag holding 3 red and 2 blue counters and draw two without replacement. The first pair of branches is $\frac{3}{5}$ and $\frac{2}{5}$ . If red was taken first, 2 red and 2 blue remain, so the second branches are $\frac{2}{4}$ and $\frac{2}{4}$ ; if blue was taken first, 3 red and 1 blue remain, so they are $\frac{3}{4}$ and $\frac{1}{4}$ . Multiplying along each route gives $P (R R) = \frac{3}{10}$ , $P (R B) = \frac{3}{10}$ , $P (B R) = \frac{3}{10}$ and $P (B B) = \frac{1}{10}$ , and those four probabilities add to 1, which is the check that the tree is labelled correctly.

Adding routes answers the usual questions. The probability of at least one red counter is $\frac{3}{10} + \frac{3}{10} + \frac{3}{10} = \frac{9}{10}$ , and the complement does it in one line: $1 - P (B B) = 1 - \frac{1}{10} = \frac{9}{10}$ . The probability that the two counters are the same colour is $\frac{3}{10} + \frac{1}{10} = \frac{2}{5}$ , since RR and BB are mutually exclusive routes.

The tree also answers a question asked backwards. Given that the second counter is red, what is the probability that the first was blue? Two routes end in red, RR and BR, and the given group is their total: $P (first blue ∣ second red) = \frac{\frac{3}{10}}{\frac{3}{10} + \frac{3}{10}} = \frac{1}{2}$ . The shape of every answer like this is the same: the route you want on top, and the total of all routes that satisfy the condition underneath.

Route	Calculation	Probability
RR	$\frac{3}{5} \times \frac{2}{4}$	$\frac{3}{10}$
RB	$\frac{3}{5} \times \frac{2}{4}$	$\frac{3}{10}$
BR	$\frac{2}{5} \times \frac{3}{4}$	$\frac{3}{10}$
BB	$\frac{2}{5} \times \frac{1}{4}$	$\frac{1}{10}$

A tree with the wrong second-stage denominators is the commonest way to lose a method mark in this topic. Without replacement the second fractions are over one fewer, and the numerators change with whatever has already been taken.

The exam standard: two counters from a bag

The standard 9709 question takes a bag, draws two or three counters without replacement, and asks for a probability. There are two correct routes and both earn method marks: the tree, where you multiply along the branches and add the routes you want, and the combinations count, where you count the selections that satisfy the condition over the total number of selections.

A bag holds 4 red and 6 blue counters and two are taken at random without replacement. By tree, the probability that they are different colours is $\frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$ . By combinations it is $\frac{(\binom{4}{1}) (\binom{6}{1})}{(\binom{10}{2})} = \frac{4 \times 6}{45} = \frac{24}{45} = \frac{8}{15}$ . Both routes give $\frac{8}{15}$ , which is the best possible check that the counting is right.

For three counters the tree has eight routes and the combinations count becomes the quicker of the two; for two counters the tree is usually clearer. Whichever route you take, write the fraction you are counting before simplifying it, because the mark scheme follows the count.

Questions asking for 'at least one' should be turned into complements. $P (at least one red) = 1 - P (no red)$ , and the route with no red is a single branch of the tree: $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ , so the answer is $1 - \frac{1}{3} = \frac{2}{3}$ . Counting the routes with at least one red directly needs three branches and invites an omission.

Tree route: multiply along each route, then add the routes that make up the event
Combination route: count the selections that satisfy the condition, over the total selections
Same colour, different colour and exactly one of each are all 'add the routes' questions
'At least one' is always quicker as $1 - P (none)$
If the two routes disagree, the counting is wrong - find the disagreement before simplifying either answer

Without replacement, a fraction such as

\frac{4}{10} \times \frac{6}{9}

is a probability of a specific order, not of the pair of colours. Different colours can happen in two orders, and both must be counted.

Key terms you must be able to define

Event: A collection of outcomes from a sample space. $P (A)$ is the probability that event A happens.

Sample space: The set of all possible outcomes of an experiment, written $ξ$ . Its probability is 1.

Equally likely outcomes: Outcomes where none is more probable than another, so each has the same probability. A probability is then a count of outcomes divided by the size of the sample space.

Complement: $A^{'}$ , everything in the sample space that is not in A. $P (A^{'}) = 1 - P (A)$ .

Mutually exclusive: Events that cannot both happen, so $P (A \cap B) = 0$ and $P (A \cup B) = P (A) + P (B)$ .

Exhaustive: Events that between them cover the whole sample space, so their probabilities add to 1.

Intersection: $A \cap B$ , the outcomes in both A and B. Read as 'A and B'.

Union: $A \cup B$ , the outcomes in A, in B, or in both. Read as 'A or B'.

Independent events: Events where $P (A \cap B) = P (A) \times P (B)$ ; knowing that one has happened does not change the probability of the other.

Dependent events: Events where the probability of one changes once the other is known, so $P (A \cap B) \neq P (A) \times P (B)$ .

Conditional probability: $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ , the probability of A given that B has happened. The sample space is restricted to B.

Tree diagram: A diagram of successive events in which each branch is labelled with a probability. Multiply along a route and add the routes that make up the event.

Two-way table: A table of counts classified two ways, used to read off conditional probabilities by taking the row or column total as the denominator.

With replacement: Each item is returned before the next is taken, so the probabilities never change and the draws are independent.

Without replacement: Items are not returned, so every draw changes the sample space and the draws are dependent.

At least one: One or more. Handled by the complement: $P (at least one) = 1 - P (none)$ .

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Two fair dice

Calculate the probability that two fair dice score a total of 9. The sample space is the 36 pairs of faces. A total of 9 needs (3, 6), (4, 5), (5, 4) or (6, 3), so there are 4 ways and $P (total = 9) = \frac{4}{36} = \frac{1}{9}$ . Counting the 11 totals as equally likely would give $\frac{1}{11}$ and lose the mark.

Different colours without replacement

A bag holds 4 red and 6 blue counters, and two are taken at random without replacement. Different colours can happen in two orders: $P (R B) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$ and $P (B R) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$ , so the answer is $\frac{48}{90} = \frac{8}{15}$ . The combination count agrees: $\frac{(\binom{4}{1}) (\binom{6}{1})}{(\binom{10}{2})} = \frac{24}{45} = \frac{8}{15}$ .

Independence in a table of counts

In a group of 60 students, 30 are boys and 24 walk to school, and 18 of the students who walk are boys. Then $P (boy) = 0.5$ , $P (walks) = 0.4$ and $P (boy) \times P (walks) = 0.2$ , but $P (boy \cap walks) = \frac{18}{60} = 0.3$ . The values differ, so the events are not independent: a student who walks is more likely to be a boy than a student chosen at random.

Conditional probability from a Venn diagram

Of 50 people, 18 own a cat but not a dog, 12 own both, 8 own a dog but not a cat and 12 own neither. Then $P (cat) = \frac{30}{50} = 0.6$ , $P (dog) = \frac{20}{50} = 0.4$ and $P (cat \cap dog) = \frac{12}{50} = 0.24$ . Since $0.6 \times 0.4 = 0.24$ , owning a cat and owning a dog are independent here, and the conditional agrees: $P (cat ∣ dog) = \frac{12}{20} = 0.6 = P (cat)$ .

At least one, by complement

Four fair coins are tossed. The probability of at least one head is $1 - P (no heads) = 1 - {(\frac{1}{2})}^{4} = 1 - \frac{1}{16} = \frac{15}{16}$ . Counting the routes with one, two, three and four heads directly needs four separate terms and gives the same answer, so the complement is worth looking for first.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1Two fair dice are thrown. What is

P (total = 7)

\frac{1}{6}

. There are 36 equally likely outcomes and six of them total 7.

Q2How many equally likely outcomes are there when two fair dice are thrown?

36, because

6 \times 6 = 36

. The 11 possible totals are not equally likely.

Q3What does

P (A ∣ B)

mean?

The probability that A happens given that B has happened. The sample space is restricted to B.

Q4State the formula for

P (A ∣ B)

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

, provided

P (B) > 0

Q5When do you add two probabilities?

When the events are mutually exclusive, so they cannot both happen:

P (A \cup B) = P (A) + P (B)

Q6When do you multiply two probabilities?

When the events are independent, or along a branch of a tree:

P (A \cap B) = P (A) \times P (B ∣ A)

Q7How do you test whether A and B are independent?

Compare

P (A \cap B)

with

P (A) \times P (B)

. Equal means independent, different means dependent.

Q8Can two mutually exclusive events be independent?

No, not when both probabilities are greater than 0. Exclusive gives

P (A \cap B) = 0

, while independence needs

P (A) \times P (B)

Q9How do you handle 'at least one'?

Use the complement:

P (at least one) = 1 - P (none)

. One route instead of three or four.

Q10What is the difference between drawing with and without replacement?

With replacement the probabilities never change and the draws are independent. Without replacement the second fraction is over one fewer.

Q11How do you find the probability of a route through a tree diagram?

Multiply the probabilities along the branches of that route. Then add the routes that make up the event you want.

Q12In a two-way table, which total is the denominator of a conditional probability?

The total of the group you are told about - the row or column named by the condition, never the grand total.

Q13A bag has 3 red and 2 blue counters. What is

P (both red)

for two counters taken without replacement?

\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}

. With replacement it would be

\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}

Q14How many ways are there of choosing 3 people from 9?

(\binom{9}{3}) = \frac{9!}{3! 6!} = 84

. Order does not matter, so it is a combination.

0 of 14 marked as known

Exam technique for this topic

Decide what the equally likely outcomes are before calculating anything, and write the size of the sample space down. 'The 36 equally likely outcomes of two dice' is worth a mark by itself, and it stops you treating the 11 totals as equally likely.
Ask whether the draws are with or without replacement. With replacement the probabilities never change; without replacement the second fraction is over one fewer, and its numerator depends on what was taken first.
For a conditional probability, choose the denominator before doing any arithmetic. The group named after 'given' is the bottom of the fraction, and the top counts the members of that group which also satisfy the other condition.
Never add probabilities unless the events are mutually exclusive, and never multiply unless the events are independent or the second factor is written as a conditional probability. If you are not told which, find $P (A \cap B)$ and compare it with $P (A) \times P (B)$ .
Check the answer against 0 and 1, and check that the branches leaving any point of a tree diagram add to 1. An answer greater than 1 always means an outcome has been counted twice.

Why students lose marks here

A student wroteThe probability that two fair dice score a total of 7 is

\frac{1}{11}

, because there are 11 possible totals.

Why it loses marksThe totals are not equally likely. A total of 7 happens in six ways and a total of 2 in only one, so the equally likely outcomes are the 36 pairs of faces.

The answer that scores

P (total = 7) = \frac{6}{36} = \frac{1}{6}

A student wroteTwo counters are taken without replacement, so

P (both red) = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}

Why it loses marksWithout replacement the second probability is conditional. One red counter has gone, so only 4 of the remaining 7 counters are red.

The answer that scores

P (both red) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

A student wroteA and B are mutually exclusive, so they must be independent.

Why it loses marksMutually exclusive means the events cannot both happen, so

P (A \cap B) = 0

. Independence needs

P (A \cap B) = P (A) \times P (B)

, which is not 0 when both probabilities are greater than 0.

The answer that scoresWith

P (A) = 0.3

and

P (B) = 0.4

, independence would need

P (A \cap B) = 0.12

. It is 0, so the events are dependent, and the two ideas are opposites here.

A student wrote

P (A ∣ B)

and

P (B ∣ A)

are the same thing.

Why it loses marksThe bar decides which group the sample space is restricted to. In the table of 60 students,

P (boy ∣ walks) = \frac{18}{24}

but

P (walks ∣ boy) = \frac{18}{30}

The answer that scores

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

and

P (B ∣ A) = \frac{P (A \cap B)}{P (A)}

. They agree only when

P (A) = P (B)

A student wroteFor a bag of 3 red and 2 blue counters, the probability of at least one red is

\frac{3}{10} + \frac{3}{10} = \frac{3}{5}

Why it loses marksOne of the three routes containing red has been missed, or a route counted twice: the routes are RR, RB and BR.

The answer that scores

P (at least one red) = 1 - P (B B) = 1 - \frac{1}{10} = \frac{9}{10}

Syllabus coverage

Events, complements and the sample space

Equiprobable elementary events: counting is the whole method

Counting outcomes with permutations and combinations

Mutually exclusive events and the addition rule

Independent events and the multiplication rule

Deciding whether two events are independent

Conditional probability: restricting the sample space

Tree diagrams and the reverse conditional probability

The exam standard: two counters from a bag

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here