Transcript Plan of Maths and Stats Lecture 4: Probability

Faculty of Social Sciences
Induction Block:
Maths & Statistics Lecture 4:
Probability, Randomness and
Different Types of Events
Dr Gwilym Pryce
1
Plan:
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1. Introduction
2. Randomness & Probability
3. Complementary and Disjoint Events
4. When 2 events can occur together
5. Independent events
6. Contingent events
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Birthday Prediction:
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
I predict that there are between one and
two birthdays in the class this week…
How did I do this?...
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– A/ 1 in 52 chance that it is your birthday
• 70 people in the room
• expected number of birthdays this week = 70 *
1/52 = 1.346
– Q/ Does that mean that every class of 70
has 1.346 birthdays that week?
– A/ No. It means that in the long run (ie lots
of lecture classes of size 70) the average
number of people with birthdays will =
1.346
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e.g. Birthdays in many classes:
Date
14/10/02
21/10/02
28/10/02
04/11/02
11/11/02
18/11/02
25/11/02
02/12/02
09/12/02
16/12/02
23/12/02
Class Number of birthdays
size
that week
Average
70
0
70
0
0.0000
70
0
0.0000
70
4
1.0000
70
5
1.8000
70
1
1.6667
70
0
1.4286
70
0
1.2500
70
0
1.1111
70
0
1.0000
70
2
1.0909 5
0.0000
22/12/02
20/12/02
18/12/02
16/12/02
14/12/02
12/12/02
10/12/02
08/12/02
06/12/02
04/12/02
02/12/02
30/11/02
28/11/02
26/11/02
24/11/02
22/11/02
20/11/02
18/11/02
16/11/02
14/11/02
12/11/02
10/11/02
08/11/02
06/11/02
04/11/02
02/11/02
31/10/02
29/10/02
27/10/02
25/10/02
23/10/02
21/10/02
Average number of birthdays
Running Average of Number of Birthdays
2.0000
1.8000
1.6000
1.4000
1.2000
1.0000
0.8000
0.6000
0.4000
0.2000
Date of Lecture
6

it can take a long time for the long run
average to emerge…
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1
Date of Lecture
8
21/10/72
21/10/70
21/10/68
21/10/66
21/10/64
21/10/62
21/10/60
21/10/58
21/10/56
21/10/54
21/10/52
21/10/50
21/10/48
21/10/46
21/10/44
21/10/42
21/10/40
21/10/38
21/10/36
21/10/34
21/10/32
21/10/30
21/10/28
21/10/26
21/10/24
21/10/22
21/10/20
21/10/18
21/10/16
21/10/14
21/10/12
21/10/10
21/10/08
21/10/06
21/10/04
21/10/02
Average number of birthdays
Long Run Average Number of Birthdays
1.6
1.5
1.4
1.3
1.2
1.1
Coin tossing example:
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2. Randomness and Probability

A phenomenon is random if
individual outcomes are uncertain
but there is nonetheless a regular
distribution of outcomes in a large
number of repetitions.
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The probability of any outcome of a
random variable is the proportion of
times the outcome would occur in a very
long series of repetitions.
– I.e. = long-term relative frequency =
number of times an event occurs in the
long run divided by the number of possible
outcomes.
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Probability of an event:
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Probability that a stranger’s birthday is
this week = 1/52
Probability that flipped coin is heads = 1/2
Probability of picking a red ball from a bag
of 2 red and 8 blue is…
– 0.2 or 20% or “1 in 5 chance” or “5 to 1
against”.

NB: probabilities always lie between 0
and 1.
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3. Probability of an event not
occurring:
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This is called the complement of an
event
It is calculated as:
• 1 - probability of the event occurring
• P(A) = 1 - P(A)
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Disjoint Events
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Two events A and B are disjoint if they
have no outcomes in common and so
can never occur simultaneously
• E.g. You have one die want to know the
probability of rolling a 4 or a 6.
• Answer: If the die is fair, the chance of rolling a
4 is 1/6. Chance of rolling a 6 is also 1/6.
Chance of rolling either a 4 or a 6 (can’t have
both at the same time) is 1/6 + 1/6 = 1/3.
• More generally, when A and B are disjoint
(mutually exclusive),
P(A or B) = P(A) + P(B)
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Venn diagram of disjoint events:
A
B
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4. When 2 events can occur together:

Suppose that you now have a die and a coin:
• What is the chance of getting a 4 and/or heads?
A
B
Roll a 4
Toss
heads (!)
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Probability of rolling 4 = 1/6
Probability of tossing heads = 1/2
Probability of rolling 4 and/or heads:
= 1/6 + 1/2 = 4/6
Or is it?
Haven’t we double counted the probability of
getting 4 and a heads?
So we need to deduct this probability (1/6*1/2):
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What we should really do is take
away (1/6*1/2):
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Probability of rolling 4 and/or heads:
= 1/6 + 1/2 - (1/6*1/2)
= 4/6 – 1/12
= 7/12
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To see this, suppose you only have a two sided die
(can only roll a one or a two) and a coin and want to
know:
Pr(1 or tails or both)
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We know that Pr(1) = ½ and Pr(heads) = ½
If we use the incorrect formula we get:
Pr(1 or heads or both) = ½+½ = 1
Which is clearly incorrect:
• There are 4 possible outcomes, 3 of which qualify as “1
or tails or both”:
(1,H) (1,T) (2,H) (2,T)
• I.e. Pr(1 or heads or both) = ¾

We have double counted Pr(both) = Pr(1,H),
where:
Pr(1,H) = ½ * ½ = ¼
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The correct formula is therefore:
P(A or B) = P(A) + P(B) - P(A and B)
Probability Notation:
A  B = A union B = A or B occur
A  B = A intersection B = A and B occur
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So we can re-write the above as:
P(A  B) = P(A) + P(B) - P(A  B)
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5. Probability of two Independent
events occurring
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If knowing that one event occurs does
not affect the outcome of another event,
we say those two outcomes are
independent.
And if A and B are independent, and we
know the probability of each of them
occurring, we can calculate the
probability of them both occurring
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Example: 2 sided die and coin,
find Pr(1 and H)
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Answer: ½ x ½ = ¼
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Rule: P(A  B) = P(A) x P(B)
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e.g. Tossing one coin twice:
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Suppose:
• A = 1st toss is a head
• B = 2nd toss is a head
– what is the probability of A  B?
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Answer: A and B are independent and
are not disjoint. P(A) = 0.5 and P(B) =
0.5. P (A  B) = 0.5 x 0.5 = 0.25.
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6. Probability of two contingent
events occurring
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If knowing that one event occurs does
change the probability that the other occurs,
then two events are not independent and are
said to be contingent upon each other
If events are contingent then we can say that
there is some kind of relationship between
them
So testing for contingency is one way of
testing for a relationship
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Example of contingent events:

There is a 70% chance that a child will go to
university if its parents are middle class, but
only a 10% chance if its parents are working
class. Given that there is a 60% chance of a
child’s parents being working class, what are
the chances that a child will be working class
and go to University? What proportion of
people at university will be from working
working class backgrounds?
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10% go to University
60%
Working Class
90% Not goto
University
Child
40%
70% go to University
Middle Class
30% Not goto
University
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6% of all children are working class
and end up going to University
10% go to University
Working Class
90% do not go to
University
60%
6% of all children
are working class
and go to University
54% of all children
are working class
and do not go to University
Child
70% go to University
40%
Middle Class
30% do not go to
University
28% of all children
are Middle class
and go to University
12% of all children
are Middle class
and do not go to University
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% = as percent of all children
Working class
Middle class
Go to
University
6%
28%
Do not go to
University
54%
12%
28
% at Uni from WC parents?
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Of all children, only 34% end up at
university (6% WC; 28% MC)
I.e 6 out of every 34 University students
are from WC parents:
6/34 = 17.65% of University students
are WC
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
Probability theory states that:
– if x and y are independent, then the probability of
events x and y simultaneously occurring is simply
equal to the product of the two events occurring:
Pr ob( x  y )  Pr ob( x)  Pr ob( x)
– if x and y are not independent, then:
Prob(x  y) = Prob(x)  Prob(y given that x has occurred)
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Test for independence

We can use these two rules to test whether
events are independent
– Does the distribution of observations across
possible outcomes resemble the random
distribution we would get if events were
independent?
– I.e. if we assume independence and calculate the
expected number of of cases in each category, do
these figures correspond fairly closely to the
actual distribution of outcomes found in our data?
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Example 1: Is there a relationship between social class and
education? We might test this by looking at categories in our
data of WC, MC, University, no University. Suppose we have
300 observations distributed as follows:
Working class
Go to University
Do not go to
University
Middle class
18
84
162
36
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To do the test for independence we need to
compare expected with observed.
How do we calculate ei, the expected number
of observations in category i?
– I.e. number of cases expected in i assuming that
the variables are independent
– the formula for ei is the probability of an
observation falling into category i multiplied simply
by the total number of observations.
• I.e. No contingency
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So, if UNIY or UNIN and WC or MC are
independent (i.e. assuming H0) then:
Prob(UNIY  WC) = Prob(UNIY)Prob(WC)
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so the expected number of cases for each of
the four mutually exclusive categories are as
follows:
Working class
Middle class
Go to University
P(UNIY) x P(WC) x n
P(UNIY) x P(MC) x n
Do not go to
University
P(UNIN) x P(WC) x n
P(UNIN) x P(MC) x n
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
But how do we work out:
Prob(UNIY)
and
Prob(WC)
which are needed to calcluate Prob(UNIY  WC):
Prob(UNIY  WC) = Prob(UNIY)Prob(WC)

Answer: we assume independence and so estimate
them from out data by simply dividing the total
observations by the total number in the given
category:
E.g. Prob(UNIY)
= Total no. cases UNIY  All observations
= (18 + 84) / 300 = 0.34
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Working class
Go to University
Do not go to
University
Middle class
P(UNIY) x P(WC) x n
= (no. at Uni / n)
x (no. WC/n)
xn
P(UNIY) x P(MC) x n
= (no. at Uni / n)
x (no. MC/n)
xn
P(UNIN) x P(WC) x n
= (no.not Uni / n)
x (no. WC/n)
xn
P(UNIN) x P(MC) x n
= (no. not Uni / n)
x (no. MC/n)
xn
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Working class Middle class
Go to
University
Do not go to
University
18
84
102
162
36
198
180
120
300
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Working class
Go to University
Do not go to
University
Middle class
P(UNIY) x P(WC) x n
= (102 / 300)
x (180 /300)
x 300
P(UNIY) x P(MC) x n
= (102 / 300)
x (120 /300)
x 300
P(UNIN) x P(WC) x n
= (198 / 300)
x (180 /300)
x 300
P(UNIN) x P(MC) x n
= (198 / 300)
x (120 /300)
x 300
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Expected count in each category:
Working class
Middle class
(102 / 300) x (180 /300) x 300
Go to
University = .34 x .6 x 300 = 61.2
(102 / 300) x (120 /300) x 300
= .34 x .4 x 300 = 40.8
Do not go
to
University
(198 / 300) x (120 /300) x 300
= .66 x .4 x 300 = 79.2
(198 / 300) x (180 /300) x 300
= .66 x .6 x 300 = 118.8
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So we have the actual count
(I.e. from our data set):
Working class
Go to University
Do not go to
University
Middle class
18
84
162
36
40
And the expected count
(I.e. the numbers we’d expect if we assume class &
education to be independent of each other):
Working class
Go to University
Do not go to
University
Middle class
61.2
40.8
118.8
79.2
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What does this table tell you?
Working class
Go to University
Middle class
18
84
Expected count
61.2
40.8
Do not go to University
162
36
118.8
79.2
Actual count
Actual count
Expected count
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
It tells you that if class and education were
indeed independent of each other
• I.e. the outcome of one does not affect the chances of
outcome of the other
– Then you’d expect a lot more working class
people in the data to have gone to university than
actually recorded (61 people, rather than 18)
– Conversely, you’d expect far fewer middle class
people to have gone to university (half the number
actually recorded).
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But remember, all this is based on a
sample, not the entire population…

Q/ Is this discrepancy due to sampling
variation alone or does it indicate that
we must reject the assumption of
independence?
44