Further Statistics — Discrete distributions: A-level Further Maths OCR revision notes

5.01

Arrangements and selections

Counting arrangements 5.01

Definitions

Permutation: an ordered arrangement. The number of ordered selections of $r$ objects from $n$ distinct objects is $^{n} P_{r} = \frac{n !}{( n - r )!}$ .
Combination: an unordered selection. The number of ways to choose $r$ objects from $n$ distinct objects is $^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$ .

Key results

$n$ distinct objects can be arranged in a line in $n!$ ways, and around a circle in $(n - 1)!$ ways (rotations count as the same).
$n$ objects of which $p$ are identical of one kind, $q$ of another, … can be arranged in $\frac{n !}{p ! q ! \dots}$ ways.
Objects that must be together: treat them as one block, arrange the blocks, then multiply by the arrangements within the block.
Objects that must be separated: arrange the others first, then place the separated objects in the gaps between them; or subtract the 'together' count from the total.

Method

Decide first whether order matters. If it does, count arrangements; if not, count selections.
Break a restricted problem into cases or stages and multiply the numbers of choices at each stage; add the counts for mutually exclusive cases.

Notes

Dividing by $p!$ removes the $p!$ orderings of identical objects that would otherwise be counted as different.
A sanity check: the number of arrangements must be a whole number, and smaller when restrictions are added.

Probability from counting 5.01

Key results

When all outcomes are equally likely, $P (A) = \frac{number of outcomes in A}{total number of outcomes}$ .
Selecting $r$ items from a group of $a$ of one type and $b$ of another: $P (k of the first type) = \frac{( k a ) ( r - k b )}{( r a + b )}$ .

Method

Count numerator and denominator in the same way: both ordered or both unordered.
For 'at least one', use $1 - P (none)$ .

Notes

A random arrangement makes every arrangement equally likely, so probabilities about arrangements are ratios of counts.
Probability calculations of this kind underlie the binomial coefficients in $B (n, p)$ and the exact tests used later in the course.

Worked examples

Worked example

(a) How many different arrangements are there of the letters of STATISTICS?

(b) One arrangement is chosen at random.

Find the probability that the three S's are together.

Show worked solution

(a) 10 letters with S three times, T three times and I twice:

\frac{10 !}{3 ! 3 ! 2 !} = 50400

(b) Treat SSS as one block: 8 items with T three times and I twice give:

\frac{8 !}{3 ! 2 !} = 3360

arrangements.

P = \frac{3360}{50 400} = \frac{1}{15}

Worked example

A committee of 4 is chosen at random from 6 men and 5 women.

Find the probability that it contains exactly 2 women.

Show worked solution

\frac{( 2 5 ) ( 2 6 )}{( 4 11 )} = \frac{10 \times 15}{330} = \frac{5}{11}

5.02

Discrete random variables

Probability distributions, expectation and variance 5.02

Definitions

Discrete random variable: a variable taking values from a countable set, each with a probability, with $\sum P (X = x) = 1$ .
Expectation (mean): $E (X) = \sum x P (X = x)$ , the long-run average value.
Variance: $Var (X) = E ((X - μ)^{2}) = E (X^{2}) - [E (X)]^{2}$ .

Key results

$E (g (X)) = \sum g (x) P (X = x)$ for any function $g$ ; in particular $E (X^{2}) = \sum x^{2} P (X = x)$ .
Discrete uniform on $1, 2, \dots, n$ : $E (X) = \frac{n + 1}{2}$ and $Var (X) = \frac{n ^{2} - 1}{12}$ .
The standard deviation is $Var (X)$ , in the same units as $X$ .

Method

If the distribution contains an unknown constant, find it first from $\sum P (X = x) = 1$ .
Tabulate $x$ , $P (X = x)$ , $x P (X = x)$ and $x^{2} P (X = x)$ ; sum the last two columns for $E (X)$ and $E (X^{2})$ .

Notes

$E (X^{2}) \neq = [E (X)]^{2}$ in general. Their difference is exactly the variance, which is never negative.
$E (X)$ need not be a value $X$ can take: on the sheet it is 2.7 for a variable taking whole-number values.

Worked example

$P (X = x) = k x$ for $x = 1, 2, 3, 4$ .

Find $k$ , $E (X)$ , $Var (X)$ , and the mean and variance of $3 X - 2$ .

Show worked solution

k (1 + 2 + 3 + 4) = 1

so:

k = \frac{1}{10}

E (X) = \frac{1 + 4 + 9 + 16}{10} = 3

E (X^{2}) = \frac{1 + 8 + 27 + 64}{10} = 10

so:

Var (X) = 10 - 9 = 1

Then:

E (3 X - 2) = 3 (3) - 2 = 7

and:

Var (3 X - 2) = 9 (1) = 9

5.02

The geometric distribution

The geometric distribution 5.02

Definitions

Geometric distribution $X \sim Geo (p)$ : the number of the trial on which the first success occurs, in independent trials each with success probability $p$ .

Key results

$P (X = x) = (1 - p)^{x - 1} p$ for $x = 1, 2, 3, \dots$
$P (X > x) = (1 - p)^{x}$ and $P (X \leq x) = 1 - (1 - p)^{x}$ .
$E (X) = \frac{1}{p}$ and $Var (X) = \frac{1 - p}{p ^{2}}$ .

Method

For tail probabilities use $P (X > x) = (1 - p)^{x}$ rather than summing terms: the first $x$ trials must all fail.
To find the least $n$ with $P (X \leq n) > c$ , solve $(1 - p)^{n} < 1 - c$ with logarithms, remembering that $ln (1 - p) < 0$ reverses the inequality.

Notes

The conditions are those of the binomial except that the number of trials is not fixed: independence and a constant $p$ .
Read the definition carefully. Here $X$ counts the trial of the first success, including it, so $X \geq 1$ .

Worked example

A fair die is rolled until a six appears.

Find the probability that the first six is on the fourth roll, and the probability that more than 10 rolls are needed.

Show worked solution

$X \sim Geo (\frac{1}{6})$ .

P (X = 4) = (\frac{5}{6})^{3} \frac{1}{6} = 0.0965

P (X > 10) = (\frac{5}{6})^{10} = 0.162

5.02

The Poisson distribution

The Poisson distribution 5.02

Definitions

Poisson distribution $X \sim Po (λ)$ : the number of events in a fixed interval of time or space, when events occur singly, independently and at a constant average rate.

Key results

$P (X = x) = \frac{e ^{- λ} λ ^{x}}{x !}$ for $x = 0, 1, 2, \dots$
$E (X) = Var (X) = λ$ .
If the rate is $λ$ per unit, the number in an interval of length $t$ is $Po (λ t)$ .
If $X \sim Po (λ)$ and $Y \sim Po (μ)$ are independent, $X + Y \sim Po (λ + μ)$ .

Method

Scale $λ$ to the interval in the question before calculating.
Use the calculator's Poisson cumulative function: $P (X \geq k) = 1 - P (X \leq k - 1)$ and $P (a \leq X \leq b) = P (X \leq b) - P (X \leq a - 1)$ .
To test $H_{0} : λ = λ_{0}$ , find the probability of the observed value or one more extreme under $H_{0}$ and compare it with the significance level.

Notes

A sample mean and variance that are close is evidence that a Poisson model is suitable; a variance much larger than the mean suggests clustering, so the events are not independent.
In context, justify each condition: 'singly' (no two at the same instant), 'independently', and 'at a constant rate'.

The Poisson approximation to the binomial 5.02

Key results

If $X \sim B (n, p)$ with $n$ large and $p$ small, then $X \approx Po (n p)$ .
The approximation works because the binomial variance $n p (1 - p)$ is close to its mean $n p$ when $p$ is small.

Notes

There is no continuity correction: both distributions are discrete.
With modern calculators the exact binomial is usually available. Use the approximation when the question asks for it or when $n$ is too large for the calculator.

Worked examples

Worked example

Calls reach a help desk at an average rate of 4 per hour.

Find the probability of exactly 6 calls in an hour, and of at least 2 calls in half an hour.

Show worked solution

For one hour, $X \sim Po (4)$ :

P (X = 6) = \frac{e ^{- 4} 4 ^{6}}{6 !} = 0.104

For half an hour, $Y \sim Po (2)$ :

P (Y \geq 2) = 1 - P (Y \leq 1) = 1 - e^{- 2} (1 + 2) = 0.594

Worked example

A machine has averaged 5 faults a week.

In one week after a service there are 10 faults.

Test at the 5% level whether the rate has increased.

Show worked solution

$H_{0} : λ = 5$ , $H_{1} : λ > 5$ , with $X \sim Po (5)$ under $H_{0}$ .

P (X \geq 10) = 1 - P (X \leq 9) = 0.0318 < 0.05

Reject $H_{0}$ : there is evidence at the 5% level that the fault rate has increased. (The critical region is $X \geq 10$ , since:

P (X \geq 9) = 0.0681 > 0.05

)

Worked example

1.5% of components are faulty.

Use a suitable approximation to find the probability that a batch of 200 contains at most 2 faulty components.

Show worked solution

X \sim B (200, 0.015)

with $n$ large and $p$ small, so $X \approx Po (3)$ .

P (X \leq 2) \approx e^{- 3} (1 + 3 + \frac{9}{2}) = 0.423

(The exact binomial value is 0.421.)

5.04

Linear combinations of random variables

Linear functions and combinations 5.04

Key results

$E (a X + b) = a E (X) + b$ and $Var (a X + b) = a^{2} Var (X)$ .
$E (a X + bY) = a E (X) + b E (Y)$ always.
If $X$ and $Y$ are independent, $Var (a X + bY) = a^{2} Var (X) + b^{2} Var (Y)$ .
For independent observations $X_{1}, \dots, X_{n}$ of $X$ : $Var (X_{1} + \dots + X_{n}) = n Var (X)$ , but $Var (n X) = n^{2} Var (X)$ .

Method

Decide first whether the question describes $n$ separate observations added together ( $X_{1} + \dots + X_{n}$ ) or one observation multiplied ( $n X$ ).
For a difference, $Var (X - Y) = Var (X) + Var (Y)$ : variances add even when the variables are subtracted.

Notes

The variance results need independence; the expectation results do not.
A sum of independent normal variables is normal, and a sum of independent Poisson variables is Poisson. Other families do not, in general, keep their shape.

Worked example

$X$ and $Y$ are independent with $E (X) = 5$ , $Var (X) = 4$ , $E (Y) = 3$ , $Var (Y) = 1$ .

Find $E (2 X - 3 Y)$ and $Var (2 X - 3 Y)$ .

Then, for independent $A \sim Po (2.5)$ and $B \sim Po (1.5)$ , find $P (A + B \leq 3)$ .

Show worked solution

E (2 X - 3 Y) = 10 - 9 = 1

Var (2 X - 3 Y) = 4 (4) + 9 (1) = 25

the variances add even though $Y$ is subtracted.

$A + B \sim Po (4)$ , so:

P (A + B \leq 3) = 0.434