Discrete Random Variables for mathematics

Discrete Random Variables
• This Chapter is on Discrete Random Variables
• We are going to learn what these are!
• Mean/ Expected value in these situations

Explanation of Discrete Random Variables
Discrete Random Variables are linked to Probability.
A Discrete Random Variable can be obtained by real-world measurement. For
example, rolling a dice.
They must always be numerical values. For example, you could toss a coin and say
‘how many heads?’, the answer being 1 or 0.
However you could not say ‘heads or tails’ as these are not numerical.
The possibilities can only be whole numbers (discrete). There can be others but
these are not the focus of the chapter.
Summary Discrete random variables allow us to calculate the expected
→
outcomes of events with given probabilities.

Notation of Discrete Random Variables
A Capital letter, such as X, will be used for the random variable, and a lower case
x for a particular value of that variable.
P(X = x) means the Probability that the Random variable is equal to a particular
value.
Rolling a Dice ‘X’
→
P(X = 5) = 1
/6
P(X > 4) = 2
/6 = 1
/3
Tossing a coin once Number of heads ‘X’
→ →
P(X = 0) = 1
/2
P(X = 1) = 1
/2

A coin is tossed 6 times and the number of heads (X) is noted. What are the
possible values of X?
→ 0, 1, 2, 3, 4, 5, 6
Which of the following are Discrete Random Variables?
→ The average height of a group of boys
→ No as height is on a continuous scale
→ The number of times a dice is rolled before a 2 appears
→ Yes, it is numerical and comes from an experiment
→ The number of months in a year
→ No as it is fixed and therefore not random
8A

You can draw up a table to show the Probability Distribution of a discrete
Random Variable. This should be something you always do first if you are not
given it in the question.
A fair dice is rolled. Show the Probability of getting any number as a Probability
Distribution.
8A
1
/6
1
/6
1
/6
1
/6
1
/6
1
/6
P(X = x)
6
5
4
3
2
1
x
→ P(X = x) = 1
/6 for x = 1, 2, 3, 4, 5, 6
This is the Probability
Function. It summarises
the data in the table.

Three fair coins are tossed. The number of
heads is counted.
a) Draw the sample space for this experiment.
→ This shows all possibilities
b) Show this as a Probability Distribution
→ The table summarises the Probabilities
H
H H
H
H
H
T
T H
T
H T
H
T
T
H
H T
T
T
T
H
T T
a)
b)
1
/8
3
/8
3
/8
1
/8
P(X = x)
3
2
1
0
No. Heads, x
These
Probabilities
will always
add up to 1

You will need to be able to calculate missing
values, based on the Probabilities adding up to
1.
a) Find the value of k in the table opposite
b) Complete the missing values in the table,
based on the value of k.
8A
3k
0.2
0.1
k
0.2
P(X = x)
5
4
3
2
1
x
0.2 + k + 0.1 + 0.2 + 3k = 1
4k + 0.5 = 1
4k = 0.5
k = 0.125
Group
together
like terms
- 0.5
÷ 4
0.375
0.2
0.1
0.125
0.2
P(X = x)
5
4
3
2
1
x

You will need to be able to calculate
missing values, based on the
Probabilities adding up to 1.
A tetrahedral (4 sided) dice is
numbered 1, 2, 3 and 4.
The probability of it landing on a given
side is k
/x, where k is constant.
a) Draw the Probability Distribution of
P(X = x), in terms of k.
b) Calculate the value of k
8A
k
/4
k
/3
k
/2
k
/1
P(X = x)
4
3
2
1
x
Make the
denominator
s equal
Now you can
group them
x 12
÷ 25

You will need to be able to calculate missing
values, based on the Probabilities adding up
to 1.
A tetrahedral (4 sided) dice is numbered 1, 2,
3 and 4.
The probability of it landing on a given side is
k
/x, where k is constant.
a) Draw the Probability Distribution of P(X =
x), in terms of k.
b) Calculate the value of k
c) Draw the finished Probability Distribution
8A
k
/4
k
/3
k
/2
k
/1
P(X = x)
4
3
2
1
x
3
/25
4
/25
6
/25
12
/25
P(X = x)
4
3
2
1
x
Half of
12
/25
12
/25 ÷ 312
/25 ÷ 4

Probability of multiple values
A discrete random variable X has the
Probability Distribution to the right
Calculate:
a)
b)
c)
d)
8B
x 1 2 3 4 5 6
P(X = x) 0.1 0.2 0.3 0.25 0.1 0.05
‘Probability X is bigger
than 1 and less than 5’
than or equal to 2 and
less than or equal to 4’
than 3 and less than or
equal to 6’
‘Probability X is less
than 3’

Probability of multiple values
Two fair coins are tossed. X is the
number of heads showing on the coins.
Draw up a sample space and then a
Probability Distribution table.
HH HT TH TT
0.25
0.5
0.25
P(X = x)
2
1
0
No.
heads, x
The Possibilities

Calculating the Expected value
The theoretical mean, μ, of a discrete random variable X is found by multiplying
each possible value of X by its probability, and then adding these products
together:
The expected value is not necessarily a value which is possible
It is effectively the mean of the distribution (this will become clearer as we do
some questions)
Notation
‘The sum of (x multiplied by
the probability of x)’
‘The expected value of x’
𝐸(𝑋)=∑
𝑖
𝑥𝑖 ×P(𝑋=𝑥𝑖)=∑
𝑖
𝑥𝑖 𝑝𝑖

Example: The probability distribution of a random variable X is:
x –2 –1 0 1
P(X = x) 0.2 0.2 0.2 0.4
E[X] = (–2 × 0.2) + (–1 × 0.2) + (0 × 0.2) + (1 × 0.4) = –0.2

The random variable X has the
following probability
distribution.
a) Given that E(X) = 3, write
down 2 equations involving p
and q.
x 1 2 3 4 5
p(x) 0.1 p 0.3 q 0.2
All the probabilities
add up to 1
Group
together the
numbers
Subtract 0.6
8C

distribution.
and q.
x 1 2 3 4 5
p(x) 0.1 p 0.3 q 0.2
Work out
each
bracket
Group
numbers
Subtract 2
8C

distribution.
and q.
b) Use your equations to find
the values of p and q.
0.2
q
0.3
p
0.1
p(x)
5
4
3
2
1
x
1
2
3
4
x 2
3
4 -
0.3 0.1
8C

Discrete Random Variables for mathematics

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