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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Lecture 21
Random Variables: Discrete &
Continuous
• A random variable is a variable whose value is
determines by the outcome of a random experiment.
• Random variables can be divided into two classes,
discrete and continuous
Discrete Random Variables
• Consider tossing a coin 3 times.
• There are 8 possible outcomes, each of which are
equally likely.
• Let X be the number of heads obtained in the
experiment.
• X depends on the outcome of the experiment and
therefore is a random variable.
• X can take one of 4 values, 0, 1, 2, 3, therefore X is
a discrete random variable
• A random variable is discrete if it can only take a
finite number of integer values or there are infintely
many values that can be labelled by an integer.
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
• A probability function can be defined to assign
probabilities to all the distinct values that X can
take, such that


 p ifx = x . 
i
i
f (x) = p(X = x) =
(19.1)
 0 otherwise. 
• The function f describes the distribution of the
random variable X.
• It is required that all probabilities sum to 1 therefore,
n
X
f (xi ) = 1,
i=1
where n may be infinite.
◦This is described as a normalisation
condition and must be satisfied by any probability
function.
• The Cumulative Probability Function can also be
useful in quantifying the probability of X taking
different values.
X
F (xi ) = p(X ≤ xi ) =
f (xi )
xi ≤x
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Example 1:
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• The cumulative proabability function F (x), may be
shown to take the values:
• Consider the tossing of a coin 3 times. Remember
the sample space has 8 different members each of
which are equally likely.
1
8
1
4
p(X ≤ 1) = F (1) = =
8
2
7
p(X ≤ 2) = F (2) =
8
8
p(X ≤ 3) = F (3) = = 1
8
p(X ≤ 0) = F (0) =
• HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
• If X is the number of heads obtained the following
are possible values for X,
{x1 , x2 , x3 , x4 } = {0, 1, 2, 3}.
• The probability of X taking each value may be
calculated by counting the number of points in the
sample space that correspond to that value.
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
f(i)
F(i)
4/8
8/8
7/8
3/8
6/8
5/8
• I.e.
1
p(X = 0) = p1 = f (0) =
8
3
p(X = 1) = p2 = f (1) =
8
3
p(X = 2) = p3 = f (2) =
8
1
p(X = 3) = p4 = f (3) =
8
• Note that
p1 + p2 + p3 + p4 = f (0) + f (1) + +f (2) + f (3) = 1
2/8
4/8
3/8
1/8
2/8
1/8
0
0
0
1
2
Number of heads
3
0
1
2
Number of heads
3
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Example 2:
The Binomial Distribution
• A bag contains 7 red balls and 3 white balls.
• 3 balls are drawn at random WITHOUT replacement.
• Find the probability function for the number of red
balls drawn.
Solution:
• The Binomial Distribution describes the processes
that consist of a number of independent, identical,
trials which have only two possible outcomes.
• For example, success and failure, where the
probability of success, p(success) = 1 − p(f ailure)
and vice versa.
Example:
• Let R be the number of red balls drawn.
• Consider a production line of children’s toys. The
toys have a probability p of being faulty.
• Then,
p(R = 0) = f (0)
=
p(R = 1) = f (1)
=
p(R = 2) = f (2)
=
p(R = 3) = f (3)
=
2 1
3
× ×
10 9 8
3
3
×
C1 ×
10
3
3
×
C2 ×
10
7
6 5
× ×
10 9 8
1
120
2 7
× =
9 8
7 6
× =
9 8
7
=
24
• A sample of n toys is taken from the line. What is the
probability that exactly r are found to be faulty?
=
7
40
21
40
• Consider the possible outcomes for the random
experiment of drawing n toys. Each outcome is a
different sequence of faulty and good toys.
• The probability of a toy being good is (1 − p),
therefore if r toys are faulty, n − r toys are good and
the probability is
pr (1 − p)n−r .
• Check that
3
X
i=0
f (i) = 1
• The number of ways that this selection could happen
is
number of ways of selecting n from r, i.e.
 the 
n

r
.
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
• Therefore, the probability of r faulty toys out of n is:


n
 pr (1 − p)n−r
p(r faulty out of n) = 
r
• This defines the probability function f (r) for the
random variable F , the number of faulty toys (out of
n sampled).
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Example 3:
• If a single six-sided die is rolled 4 times, what is the
probability that a 5 is thrown exactly 2 times?
Solution:
• The number of trials, n = 4
• The possible values for B are {0, 1, ....., n}.
• The random variable, X= number of 5’s thrown.
• Note that
• The probability of throwing a 5 (success), p = 1/6
n
X
p(r faulty out of n)
=
r=0
n
X
• Therefore,
f (r)
r=0
=
n
X
r=0


n
r

 pr (1 − p)n−r
= [p + (1 − p)]n
= 1n
= 1
• The distribution of F is called the binomial
distribution, B(n, p), specified by the two
parameters n (the number in the sample) and p (the
probability of one of the two possible types).
f (x) = p(X = x) =n Cr px (1 − p)n−x (1)
4!
p(X = 2) =
2!(4 − 2)!
2 (4−2)
5
1
= 0.12
6
6
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
Example 4:
• If two six-sided dice are rolled 3 times
(a) What is the probability that a total of 12 is
thrown exactly twice?
(b) What is the probability that a total of 7 is
thrown exactly 3 times?
Solution:
(a)
• The number of trials, n = 3
Graphically:
• Return to our toys on their production line:
• Calculate the probability of n good toys in N
produced. What happens to the distribution as N
increases?
• If p(good) = 0.5, N = 5, N = 10 and N = 50.
• The random variable, X= number of 12’s thrown.
• The probability of throwing a total of 12 (success),
p = 1/36
• Therefore,
3!
p(X = 2) =
2!(3 − 2)!
1
36
2 35
36
(3−2)
= 0.002
(b)
• The number of trials, n = 3
• The random variable, X= number of 7’s thrown.
• The probability of throwing a total of 7 (success),
p = 6/36
• Therefore,
3!
p(X = 3) =
3!(3 − 3)!
3 (3−3)
5
1
= 0.005
6
6
• Example for N = 5:
5!
0.50 0.55
p(X = 0) =
0!(5 − 0)!
5!
p(X = 1) =
0.51 0.54
1!(5 − 1)!
5!
p(X = 2) =
0.52 0.53
2!(5 − 2)!
5!
p(X = 3) =
0.53 0.52
3!(5 − 3)!
5!
0.54 0.51
p(X = 4) =
4!(5 − 4)!
5!
p(X = 5) =
0.55 0.50
5!(5 − 5)!
= 0.03125
= 0.15625
= 0.3125
= 0.3125
= 0.15625
= 0.03125
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
• Let p(good) = 0.3, N = 5, N = 10 and N = 50.
• Example for N = 5:
5!
p(X = 0) =
0.30 0.75
0!(5 − 0)!
5!
0.31 0.74
p(X = 1) =
1!(5 − 1)!
5!
p(X = 2) =
0.32 0.73
2!(5 − 2)!
5!
0.33 0.72
p(X = 3) =
3!(5 − 3)!
5!
p(X = 4) =
0.34 0.71
4!(5 − 4)!
5!
0.35 0.70
p(X = 5) =
5!(5 − 5)!
= 0.16807
= 0.0.36015
= 0.0.3087
= 0.1323
= 0.02835
= 0.00243
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NST1A Maths Course A, Lisa Jardine-Wright, Michaelmas 2006
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