A LevelFurther MathsEdexcelFurther Statistics 1Revision NotesDiscrete Probability DistributionsDiscrete Probability DistributionsProblem Solving with DRVs

Problem Solving with DRVs (Edexcel A Level Further Maths: Further Statistics 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 19 June 2025

Problem solving with DRVs

How do I find DRVs from worded contexts?

Introduce a capital letter, $X$ , to represent numerical values
- Make sure you know exactly what $X$ represents
List all the values $X$ can take
Calculate the probability for each value of $X$
Put this information into a table
$x$
...
$P (X = x)$
...
If asked to comment on results
- Find $E (X)$ to talk about the average result
- Find $Var (X)$ to talk about the spread of results

What if they introduce a new independent variable?

Sometimes questions have two independent DRVs, $X$ and $Y$
- e.g. $X$ may be a coin and $Y$ may be a 3-sided spinner
Construct their tables
$x$
2
4
$P (X = x)$
0.25
0.75
$y$
0
2
4
$P (Y = y)$
0.1
0.2
0.7
To find $P (X + Y = 4)$
- List possibilities
  - $X = 2, Y = 2$ or $X = 4, Y = 0$
- Multiply probabilities then add
  - 0.25 × 0.2 + 0.75 × 0.1
To find $E (X + Y)$
- Either construct a full table for all possible values that $X + Y$ can take
  - There may be a lot!
  - Then work out expectation from the table
- Or use the formula $E (X + Y) = E (X) + E (Y)$
  - Find $E (X)$ and $E (Y)$ from their tables above
Note that, if $X$ and $Y$ are independent
- Then $E (X Y) = E (X) E (Y)$
- The expectation of a product is the product of the expectation
These ideas can be applied when repeatedly using the same distribution, $X$
- For example, flipping a coin, then flipping it again
- Create independent random variables $X_{1}, X_{2}, . . .$

What if they introduce a new dependent variable?

Sometimes questions introduce a new variable, $Y$ , which depends on an old variable, $X$
$x$
-3
1
2
3
$P (X = x)$
0.25
0.25
0.25
0.25
- Let $Y = 9 - X^{2}$
  - List values of $y$
  - They have the same probabilities as $X$ (as $Y$ depends on $X$ )
  - $y$
    0
    8
    5
    0
    $P (Y = y)$
    0.25
    0.25
    0.25
    0.25
    Simplify (reorder and group values together)
  $y$
  0
  5
  8
  $P (Y = y)$
  0.5
  0.25
  0.25
You can use $E (X + Y) = E (X) + E (Y)$
- calculate $E (X)$ and $E (Y)$ from their tables
You cannot use $E (X Y) = E (X) E (Y)$ if $Y$ depends on $X$
- To find $E (X Y)$ draw the table for all possible values of $X Y$
  $x y$
  -3 × 0
  1 × 8
  2 × 5
  3 × 0
  $P (X = x)$
  0.25
  0.25
  0.25
  0.25
  - Probabilities are the same as $X$ (as $Y$ depends on $X$ )
  - Then calculate expectation from this table
- Alternatively, substitute in $Y$
  - $X Y = X (9 - X^{2}) = 9 X - X^{3}$
  - So $E (9 X - X^{3}) = 9 E (X) - E (X^{3})$
  - Find $E (X)$ from the table of $X$
  - Find $E (X^{3})$ from the table of $X$ (by cubing the $x$ values)

What if there is conditional probability?

Learn the conditional probability formula
- $P (A | B) = \frac{P (A \cap B)}{P (B)}$
- For example, for integer values, the probability that $X > 5$ given that $X \leq 10$
  - $P (X > 5 | X \leq 10) = \frac{P (6 \leq X \leq 10)}{P (X \leq 10)}$

Examiner Tips and Tricks

Always draw out tables of values for each DRV - they really help in the exam!

Worked Example

In a game, you can earn 2, 5 or 6 points. There is a 50% chance of earning 6 points and an equal chance of earning either 2 or 5 points.

a) Find the probability of earning more than 9 points when playing the game twice.

The amount of money (£) won in a game is found by multiplying the number of points, $X$ , by the variable $Y$ , where $Y = 3$ if $X < 5$ or $Y = X - 3$ if $X \geq 5$ .

b) Find the expected amount of money won per game.

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

$x$	...
$P (X = x)$	...

$x$	2	4
$P (X = x)$	0.25	0.75

$y$	0	2	4
$P (Y = y)$	0.1	0.2	0.7

$x$	-3	1	2	3
$P (X = x)$	0.25	0.25	0.25	0.25