Discrete Random Variables

Syllabus coverage

Draw up a probability distribution table relating to a given situation involving a discrete random variable X, and calculate $E (X)$ and $V a r (X)$
Use the formulae for probabilities for the binomial and geometric distributions, and recognise practical situations where these distributions are suitable models, including the notations $B (n, p)$ and $G e o (p)$
Use the formulae for the expectation and variance of the binomial distribution and for the expectation of the geometric distribution (proofs of the formulae are not required)

Part of the AS & A Level Maths syllabus 9709.

One discrete distribution in three forms: a probability table, the binomial

B (10, 0.3)

with mean

n p = 3

, and the geometric

G e o (0.2)

with mean

1 / p = 5

. The probabilities in the table add to 1.

What a discrete random variable is

A random variable is a numerical value that depends on the outcome of an experiment. A discrete random variable takes values that can be listed one by one, and each of those values has a probability attached to it. If $X$ is the number of heads when three coins are tossed, then $X$ can be 0, 1, 2 or 3, and each of those four values has its own probability.

The method never changes: write the possible values down in order, then attach a probability to each one. If a value can happen it must appear in the list, each value appears once, and the probabilities must account for the whole experiment.

Variables that are measured rather than counted - a waiting time, a height, the mass of a bag of flour - are continuous, and they belong to the normal distribution later in the course.

Situation	The random variable	Possible values
A fair die is thrown	$X$ = the score shown	1, 2, 3, 4, 5, 6
Three fair coins are tossed	$X$ = the number of heads	0, 1, 2, 3
A box holds 20 components	$X$ = the number of faulty components in the box	0, 1, 2, ..., 20
A counter is drawn from a bag and returned	$X$ = 1 if it is red and 0 otherwise	0, 1

Ask whether the variable is counting something or measuring something. 'The number of...' is discrete; 'the time taken', 'the mass of...' are continuous and are not modelled by a distribution table.

Probability distribution tables

A probability distribution of a discrete random variable is the list of its values together with their probabilities. It is written as a table whose top row is the values $x$ and whose bottom row is $P (X = x)$ . Sometimes the probabilities are given by a formula instead, such as $P (X = x) = k x$ , and then the table is built from the formula.

Every distribution must satisfy two conditions. Each probability lies between 0 and 1, and the probabilities add up to exactly 1, because between them the values cover everything that can happen. Those two conditions are the whole of the theory, and the second one is how an unknown constant in a distribution is found.

So if a variable takes the values 1, 2, 3 and 4 with $P (X = x) = k x$ , then $k + 2 k + 3 k + 4 k = 10 k = 1$ , giving $k = 0.1$ . Only then can any probability be worked out: $P (X = 3) = 0.3$ and $P (X \geq 3) = 0.3 + 0.4 = 0.7$ .

The distribution of the total when two fair dice are thrown is worth knowing by heart, because it shows how unevenly a distribution can be spread. There are 36 equally likely outcomes and the totals run from 2 to 12, with the counts 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1.

$x$	2	3	4	5	6	7	8	9	10	11	12
$P (X = x)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

Fractions and decimals are both fine, but do not mix them in one table. A table whose probabilities add to 0.9 is missing a value; one that adds to 1.1 has counted a value twice.

Expectation E(X)

The expectation of a discrete random variable is $E (X) = \sum x P (X = x)$ : multiply every value by its probability and add the results. It is the average value of $X$ over a very large number of repetitions of the experiment, and it is the point at which the distribution would balance.

For a fair die, $E (X) = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5$ . Notice that 3.5 is not a face of the die. An expectation does not have to be a value the variable can actually take, and it does not have to be a whole number: it is an average, not an outcome.

For the total when two dice are thrown, $E (X) = 7$ without any arithmetic at all, because the distribution is symmetrical about 7. Symmetry is worth looking for before you start multiplying, and it also gives you a free check on a sum you have already done.

The probabilities act as weights, so values with larger probabilities pull the mean towards them: in the two-dice table the value 7 carries the largest probability and the mean sits exactly there.

$x$	$P (X = x)$	$x P (X = x)$
1	$\frac{1}{6}$	$\frac{1}{6}$
2	$\frac{1}{6}$	$\frac{2}{6}$
3	$\frac{1}{6}$	$\frac{3}{6}$
4	$\frac{1}{6}$	$\frac{4}{6}$
5	$\frac{1}{6}$	$\frac{5}{6}$
6	$\frac{1}{6}$	$\frac{6}{6}$

An expectation is a mean, so it is a number and never a percentage or a probability. If your

E (X)

lies outside the range of the possible values, something has been multiplied wrongly.

Variance and standard deviation

The variance measures how spread out the values are, and the formula list gives it as $V a r (X) = \sum x^{2} P (X = x) - {E (X)}^{2}$ , usually written $V a r (X) = E (X^{2}) - {E (X)}^{2}$ . There are two separate sums: one for the mean of the values, $E (X)$ , and one for the mean of the squares, $E (X^{2})$ .

For a fair die, $E (X^{2}) = \frac{1}{6} (1 + 4 + 9 + 16 + 25 + 36) = \frac{91}{6} = 15.1667$ , so $V a r (X) = \frac{91}{6} - {(\frac{7}{2})}^{2} = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} = 2.9167$ , correct to 4 decimal places. The standard deviation is the square root of the variance, $\sqrt{2.9167} = 1.71$ , and it is measured in the same units as $X$ .

A small variance means the values cluster near the mean; a large one means they are spread wide. Because every squared term is positive, a variance can never be negative - if you get a negative value you have subtracted the square of the mean from too small an $E (X^{2})$ , which usually means the two sums have been swapped.

Working in exact fractions and rounding only at the end keeps the last figure right. $\frac{91}{6} - \frac{49}{4}$ is exactly $\frac{35}{12}$ ; writing 15.17 and 12.25 and subtracting gives 2.92, which is close but not the value the mark scheme prints.

$x$	$P (X = x)$	$x P (X = x)$	$x^{2} P (X = x)$
1	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$
2	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{4}{6}$
3	$\frac{1}{6}$	$\frac{3}{6}$	$\frac{9}{6}$
4	$\frac{1}{6}$	$\frac{4}{6}$	$\frac{16}{6}$
5	$\frac{1}{6}$	$\frac{5}{6}$	$\frac{25}{6}$
6	$\frac{1}{6}$	$\frac{6}{6}$	$\frac{36}{6}$
Total	1	$\frac{7}{2}$	$\frac{91}{6}$

Subtract the square of the mean, not the mean itself.

\frac{91}{6} - {(\frac{7}{2})}^{2} = 2.9167

, while subtracting 3.5 gives 11.67 - a spread wider than the whole die, which is the giveaway that the wrong formula has been used.

The binomial distribution B(n, p)

The binomial distribution counts the number of successes in a fixed number of independent trials, where every trial has the same probability of success. Four conditions have to hold: a fixed number $n$ of trials, independent trials, two possible outcomes per trial, and the same probability $p$ of success every time. When they do, we write $X ~ B (n, p)$ , where $n$ is the number of trials and $p$ is the probability of success on one trial.

The probability formula is $P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ for $r = 0, 1, 2, \dots, n$ . The two parts do different jobs. The term $p^{r} (1 - p)^{n - r}$ is the probability of one particular order of $r$ successes and $n - r$ failures, and the coefficient $(\binom{n}{r})$ counts how many different orders there are. Leaving out the coefficient is the commonest way to lose the accuracy mark.

For $X ~ B (10, 0.3)$ , $P (X = 3) = (\binom{10}{3}) (0.3)^{3} (0.7)^{7} = 120 \times 0.027 \times 0.0823543 = 0.2668$ , correct to 4 decimal places. Reading it back: 120 orders of three successes among ten trials, each order having probability 0.00222357.

Both numbers in the notation have a meaning that the question tests. $B (20, 0.15)$ means 20 trials with a probability of 0.15 of success; $B (0.15, 20)$ is not the same thing at all, and 20 is not a probability.

Situation	Binomial?	Why
The number of sixes in 20 throws of a fair die	Yes	$n = 20$ fixed, throws independent, $p = \frac{1}{6}$ constant
The number of questions a student guesses correctly in a test of 12 with 4 options each	Yes	$X ~ B (12, 0.25)$ if the answers are all guesses
The number of red counters in 5 counters drawn without replacement	No	the probability changes after each draw, so the trials are not independent
The number of throws until the first six appears	No	the number of trials is not fixed - that is the geometric distribution

Write

X

down in words before you use the formula: 'Let

X

be the number of seeds that germinate, so

X ~ B (200, 0.85)

'. That sentence shows the examiner which model you are using and is worth a mark on its own.

Binomial probabilities for at least, at most and exactly

Most questions ask for a range rather than a single value, and the answer is a sum of the terms the formula gives. There are no binomial tables in the 9709 formula booklet, so the terms are added by hand - which is why the questions keep the numbers small.

'At least one' is the case worth memorising: $P (X \geq 1) = 1 - P (X = 0) = 1 - (1 - p)^{n}$ , because the only way to fail is for every trial to fail. With $X ~ B (8, 0.25)$ , $P (X \geq 1) = 1 - {0.75}^{8} = 1 - 0.1001129 = 0.8999$ , one line instead of eight terms.

'At least two' needs the terms for 0 and for 1: $P (X \geq 2) = 1 - P (X \leq 1)$ . With $X ~ B (20, 0.15)$ , $P (X \leq 1) = {0.85}^{20} + 20 (0.15) (0.85)^{19} = 0.03876 + 0.13680 = 0.17556$ , so $P (X \geq 2) = 1 - 0.17556 = 0.8244$ to 4 decimal places.

'At most' and 'fewer than' go the other way. $P (X \leq 2)$ is the sum of the terms for 0, 1 and 2 with no subtraction, while $P (X < 2)$ is the sum of the terms for 0 and 1 only, because the value 2 is excluded by the strict inequality.

$P (X = 0) = (1 - p)^{n}$ , and $P (X = n) = p^{n}$
$P (X \geq 1) = 1 - (1 - p)^{n}$ , the quickest of all the range formulae
$P (X \geq 2) = 1 - P (X = 0) - P (X = 1)$
$P (X > 2) = 1 - P (X = 0) - P (X = 1) - P (X = 2)$ , one term more than $P (X \geq 2)$
Every range probability is a sum of terms of the form $(\binom{n}{r}) p^{r} (1 - p)^{n - r}$

'At least 2' is not 'more than 2'.

P (X \geq 2) = 1 - P (X \leq 1)

leaves the term for 1 out of the sum, while

P (X > 2) = 1 - P (X \leq 2)

includes it. Reading the inequality carefully is worth more than any shortcut.

The mean and variance of the binomial distribution

For $X ~ B (n, p)$ the formula list gives $E (X) = n p$ and $V a r (X) = n p (1 - p)$ , with standard deviation $\sqrt{n p (1 - p)}$ . Proofs are not required, so these are substituted into directly - but the meaning is worth holding on to: on average a proportion $p$ of the trials succeed.

For $X ~ B (20, 0.15)$ , $E (X) = 20 \times 0.15 = 3$ prizes on average and $V a r (X) = 20 \times 0.15 \times 0.85 = 2.55$ , so the standard deviation is $\sqrt{2.55} = 1.60$ . The mean and the variance are close here because $p$ is small; when $p = 0.5$ the variance is at its largest for a given $n$ , and it falls away towards 0 as $p$ approaches 0 or 1.

The two formulae are also used backwards, which is where marks are usually lost. If $X ~ B (8, p)$ and $E (X) = 2.4$ , then $8 p = 2.4$ , so $p = 0.3$ , and the variance follows at once: $8 \times 0.3 \times 0.7 = 1.68$ . If instead the variance is given as 2.1 with $p = 0.3$ , then $n \times 0.3 \times 0.7 = 2.1$ , so $0.21 n = 2.1$ and $n = 10$ .

Quantity	Formula	For $B (20, 0.15)$
Mean	$E (X) = n p$	$20 \times 0.15 = 3$
Variance	$V a r (X) = n p (1 - p)$	$20 \times 0.15 \times 0.85 = 2.55$
Standard deviation	$\sqrt{n p (1 - p)}$	$\sqrt{2.55} = 1.60$

E (X) = n p

is an average count of successes, so it can be larger than 1 and need not be a whole number. Do not report it as a probability or as a percentage.

The geometric distribution Geo(p)

The geometric distribution counts the number of trials up to and including the first success. The trials are independent, the probability of success is the same $p$ on every trial, and the experiment stops as soon as a success occurs. We write $X ~ G e o (p)$ , and the formula list gives $P (X = r) = p (1 - p)^{r - 1}$ for $r = 1, 2, 3, \dots$ together with $E (X) = \frac{1}{p}$ .

Read the formula as a story: $r - 1$ failures, each with probability $1 - p$ , followed by one success with probability $p$ . The smallest possible value of $X$ is 1, because the first trial could succeed, and there is no upper limit, because the first success might be a long way off.

For $X ~ G e o (0.2)$ , $P (X = 3) = 0.2 \times {0.8}^{2} = 0.128$ and $E (X) = \frac{1}{0.2} = 5$ trials. The mean is worth reading as a sentence: with a probability of 0.2 of success, you expect to wait five trials.

A related probability comes up often: $P (X > r) = (1 - p)^{r}$ , because $X > r$ means the first $r$ trials all fail. For the same distribution, $P (X > 3) = {0.8}^{3} = 0.512$ , and $P (X \geq 3) = {0.8}^{2} = 0.64$ , since $X \geq 3$ allows the third trial to succeed. The variance of the geometric distribution is not required at 9709 - the formula list gives only its mean - so never reach for a geometric variance formula.

$r$	$P (X = r)$ for $G e o (0.2)$
1	0.2
2	$0.2 \times 0.8 = 0.16$
3	$0.2 \times {0.8}^{2} = 0.128$
4	$0.2 \times {0.8}^{3} = 0.1024$
5	$0.2 \times {0.8}^{4} = 0.08192$
6	$0.2 \times {0.8}^{5} = 0.065536$

The geometric variable counts the trial on which the first success happens, so its smallest value is 1 and never 0. A table that starts at 0 has been built for the wrong variable.

Choosing the right model

'Recognise practical situations where these distributions are suitable models' is an objective in its own right, so the choice of model is often worth a mark. The question to ask first is whether the number of trials is fixed. If it is fixed, and the trials are independent with a constant probability of success, the model is binomial. If the trials continue until the first success, the model is geometric.

The condition that is most often broken is independence. Drawing counters from a bag without replacement changes the probability after every draw, so a binomial model is not suitable, however much the question looks like one. Sampling without replacement from a very large population is the exception: the probability changes so little that the binomial is used as a good approximation, which is why 'a large population' appears in the wording of those questions.

Two more situations do not fit either model and are outside this syllabus: counting successes in a sample from a small population without replacement, and counting events that occur independently in time or space. If a question describes one of those, it will give you the distribution table instead.

Whatever the model, the working starts the same way: define the variable in words, name the distribution with its parameters, and then substitute into the formula.

The situation	The model	The parameters
A fixed number of independent trials, counting the successes	$X ~ B (n, p)$	$n$ trials, probability $p$ of success on each
Independent trials repeated until the first success, counting the trials	$X ~ G e o (p)$	probability $p$ of success on each trial
Sampling without replacement from a small population	neither - the trials are not independent	use the probability distribution table instead
Sampling from a large population, or from a small sample of a large one	$X ~ B (n, p)$ as a good approximation	the probability is treated as constant

Binomial or geometric comes down to one question: is the number of trials decided in advance? Twenty throws of a die is binomial; throws until the first six is geometric. Both count successes and both need independent trials with a constant probability.

Key terms you must be able to define

Random variable: A numerical value determined by the outcome of an experiment.

Discrete random variable: A random variable whose possible values can be listed one by one, each with its own probability.

Probability distribution: The table or formula that gives the probability of each value of a discrete random variable. The probabilities are all between 0 and 1 and add up to 1.

Expectation $E (X)$: $E (X) = \sum x P (X = x)$ , the mean value of X over a very large number of repetitions. It need not be a value X can take.

$E (X^{2})$: $\sum x^{2} P (X = x)$ , the mean of the squares of the values, used in the variance formula.

Variance $V a r (X)$: $V a r (X) = E (X^{2}) - {E (X)}^{2}$ , a measure of how spread out the distribution is. It is never negative.

Standard deviation: The square root of the variance, in the same units as X. For a fair die it is $\sqrt{\frac{35}{12}} = 1.71$ .

Trial: One performance of the experiment: one throw of a die, one toss of a coin, one item inspected.

Success: The outcome a trial is being counted for. The other outcome is a failure, and its probability is $1 - p$ .

Binomial distribution: $X ~ B (n, p)$ , the number of successes in $n$ independent trials with constant probability $p$ . $P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ , $E (X) = n p$ and $V a r (X) = n p (1 - p)$ .

Geometric distribution: $X ~ G e o (p)$ , the number of trials up to and including the first success. $P (X = r) = p (1 - p)^{r - 1}$ for $r = 1, 2, 3, \dots$ and $E (X) = \frac{1}{p}$ .

Independent trials: Trials where the outcome of one does not change the probability of the next. Both the binomial and the geometric models need them.

Cumulative probability: A probability of the form $P (X \leq r)$ , found by adding the individual probabilities from the smallest value up to $r$ .

Complement of an event: In this topic, the quickest route to an 'at least' probability: $P (X \geq r) = 1 - P (X \leq r - 1)$ .

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.

A distribution table built from scratch

Three fair coins are tossed and $X$ is the number of heads. The values are 0, 1, 2, 3 with probabilities $\frac{1}{8}$ , $\frac{3}{8}$ , $\frac{3}{8}$ , $\frac{1}{8}$ , which add to 1. Then $E (X) = \frac{3}{2} = 1.5$ , $E (X^{2}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = 3$ , and $V a r (X) = 3 - {1.5}^{2} = 0.75$ .

A binomial probability

A multiple-choice test has 10 questions with 4 options each, and a student guesses every answer. With $X ~ B (10, 0.25)$ , $P (X = 4) = (\binom{10}{4}) (0.25)^{4} (0.75)^{6} = 210 \times 0.00390625 \times 0.1779785 = 0.1460$ . The coefficient 210 is the number of ways the four correct guesses can be arranged among the ten questions.

A geometric probability

Five per cent of the items from a machine are defective, independently. With $X ~ G e o (0.05)$ the number of items inspected up to and including the first defective one, $P (X = 10) = 0.05 \times {0.95}^{9} = 0.05 \times 0.6302494 = 0.0315$ , and $E (X) = \frac{1}{0.05} = 20$ items.

Expectation and variance of a binomial

Two hundred seeds are sown, each germinating with probability 0.85, independently. With $X ~ B (200, 0.85)$ , $E (X) = 200 \times 0.85 = 170$ seeds and $V a r (X) = 200 \times 0.85 \times 0.15 = 25.5$ , so the standard deviation is $\sqrt{25.5} = 5.05$ seeds.

Working backwards from the mean

A random variable has the distribution $B (20, p)$ and $E (X) = 5$ . Then $20 p = 5$ , so $p = 0.25$ , and $V a r (X) = 20 \times 0.25 \times 0.75 = 3.75$ with standard deviation $\sqrt{3.75} = 1.94$ . Working backwards is the same substitution read in the other direction.

Flashcards

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

Q1What two conditions must a probability distribution table satisfy?

Every probability lies between 0 and 1, and the probabilities add up to exactly 1.

Q2State the formula for

E (X)

E (X) = \sum x P (X = x)

: multiply each value by its probability and add the results.

Q3State the formula for

V a r (X)

V a r (X) = E (X^{2}) - {E (X)}^{2}

, where

E (X^{2}) = \sum x^{2} P (X = x)

Q4For a fair six-sided die, what are

E (X)

and

V a r (X)

E (X) = 3.5

and

V a r (X) = \frac{91}{6} - 12.25 = \frac{35}{12} = 2.9167

Q5What are the four conditions for a binomial distribution?

A fixed number of trials, independent trials, two outcomes per trial, and the same probability of success in every trial.

Q6Write down

P (X = r)

for

X ~ B (n, p)

P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}

for

r = 0, 1, \dots, n

Q7What are

E (X)

and

V a r (X)

for

X ~ B (n, p)

E (X) = n p

and

V a r (X) = n p (1 - p)

Q8Write down

P (X = r)

for

X ~ G e o (p)

P (X = r) = p (1 - p)^{r - 1}

for

r = 1, 2, 3, \dots

Q9What is the mean of a geometric distribution?

E (X) = \frac{1}{p}

. The variance of the geometric distribution is not required at 9709.

Q10How do you find

P (X \geq 1)

quickly for a binomial variable?

Use the complement:

P (X \geq 1) = 1 - P (X = 0) = 1 - (1 - p)^{n}

Q11How do you find

P (X > r)

for a geometric variable?

P (X > r) = (1 - p)^{r}

, because the first

r

trials must all fail.

Q12Which distribution models the number of throws of a fair die up to and including the first six?

Geometric with

p = \frac{1}{6}

, so

E (X) = 6

throws on average.

Q13Which distribution models the number of sixes in 20 throws of a fair die?

Binomial:

X ~ B (20, \frac{1}{6})

, with

E (X) = 20 \times \frac{1}{6} = 3.33

sixes on average.

Q14Why is drawing counters without replacement not a binomial situation?

The probability of success changes after every draw, so the trials are not independent and p is not constant.

0 of 14 marked as known

Exam technique for this topic

Define the variable in words and name its distribution before calculating: 'Let $X$ be the number of seeds that germinate, so $X ~ B (200, 0.85)$ '. That sentence shows which model you are using and usually carries a mark.
Check that a probability distribution table adds to 1 before you use it. A table adding to 0.9 has a missing value, and one adding to 1.1 has counted a value twice.
Use the complement for 'at least one': $P (X \geq 1) = 1 - (1 - p)^{n}$ . It is one term instead of n terms, and it removes the risk of leaving a term out.
Substitute into $E (X) = n p$ and $V a r (X) = n p (1 - p)$ for a binomial, but remember that only the mean $\frac{1}{p}$ is available for a geometric distribution. Never use a geometric variance formula - it is not on the formula list.
Give probabilities to three significant figures unless the question asks for more, keep full accuracy in the working, and check the answer is between 0 and 1.

Why students lose marks here

A student wroteThe number of red counters in 5 counters drawn without replacement from a bag is a binomial variable.

Why it loses marksWithout replacement the probability of a red counter changes after every draw, so the trials are not independent and p is not constant. Those are two of the four conditions for a binomial model.

The answer that scoresA binomial model needs a constant probability on every trial. Here p falls from

\frac{8}{20}

\frac{7}{19}

\frac{6}{18}

, so the variable is not binomial.

A student wrote

V a r (X) = \sum x^{2} P (X = x)

Why it loses marksThat sum is

E (X^{2})

, not the variance. The variance is the mean of the squares minus the square of the mean.

The answer that scores

V a r (X) = E (X^{2}) - {E (X)}^{2}

. For a fair die,

\frac{91}{6} - 12.25 = 2.9167

A student wroteFor

X ~ G e o (0.2)

the expected number of trials is

n p

, and I used

E (X) = \frac{1}{p}

for the binomial.

Why it loses marksThe two means belong to different models.

n p

counts successes in a fixed number of trials;

\frac{1}{p}

counts the trials until the first success.

The answer that scoresFor

B (n, p)

E (X) = n p

; for

G e o (p)

E (X) = \frac{1}{p}

. With

p = 0.2

the geometric mean is 5 trials.

A student wrote

P (X \geq 2) = 1 - P (X = 2)

Why it loses marks'At least 2' excludes 0 and 1, so both of those terms have to come out of the complement - not just the term for 2.

The answer that scores

P (X \geq 2) = 1 - P (X \leq 1) = 1 - P (X = 0) - P (X = 1)

A student wroteThe distribution is

B (0.15, 20)

, with 0.15 trials and 20 successes.

Why it loses marksThe order of the two numbers in

B (n, p)

is fixed by the notation: the number of trials comes first, the probability of success second.

The answer that scores

B (20, 0.15)

n = 20

trials with

p = 0.15

. The same order applies to

G e o (p)

, which takes only the one parameter.

Syllabus coverage

What a discrete random variable is

Probability distribution tables

Expectation E(X)

Variance and standard deviation

The binomial distribution B(n, p)

Binomial probabilities for at least, at most and exactly

The mean and variance of the binomial distribution

The geometric distribution Geo(p)

Choosing the right model

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