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AS-Level Maths:
Statistics 1
for OCR MEI
S1.3 Probability
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Probability basics and notation
Contents
Probability basics and notation
Estimating probability
Addition properties
Independent events
Conditional probability
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Probability
How likely am I to live to 100?
Which team is most likely to win the FA cup?
Are interest rates likely to go up?
Am I likely to win the lottery?
Uncertainty is a feature of everyday life. Probability is an
area of maths that addresses how likely things are to
happen.
A good understanding of probability is important in many
areas of work. It is used by scientists, governments,
businesses, insurance companies, betting companies and
many others, to help them anticipate future events.
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Introduction to probability
A statistics experiment will have a number of different
outcomes. The set of all possible outcomes is called the
sample space of the experiment.
For example:
if a normal dice is thrown the sample
space would be {1, 2, 3, 4, 5, 6}.
In a general knowledge quiz with 70 questions,
the sample space for the number of questions a
person answers correctly is {0, 1, 2, …, 70}.
An event is a collection of some of the outcomes from an
experiment. For example, getting an even number on the dice
or scoring more than 40 on the quiz.
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Notation
Let A be an event arising from a statistical experiment.
The probability that A occurs is denoted P(A)
(where 0 ≤ P(A) ≤ 1).
If A is certain to happen, then P(A) = 1.
If A is impossible, then P(A) = 0.
The probability that A does not occur is denoted P(A′).
P(A′) = 1 – P(A)
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Introduction to probability
When two experiments are combined, the set of possible
outcomes can be shown in a sample space diagram.
There are 36 equally
likely outcomes.
5 of the outcomes
result in a total of 6.
P(total = 6) = 5
36
Second throw
Example: A dice is thrown twice and the scores obtained are
added together. Find the probability that the total score is 6.
6
7
8
9
10
11
12
5
6
7
8
9
10
11
4
5
6
7
8
9
10
3
4
5
6
7
8
9
2
3
4
5
6
7
8
1
2
3
4
5
6
7
1
2
5
6
3
4
First throw
This notation means “probability that the total = 6”.
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Estimating probability
Contents
Probability basics and notation
Estimating probability
Addition properties
Independent events
Conditional probability
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Estimating probability
Some probabilities are less simple. It is not always possible to
calculate how likely each outcome is.
However, the probability of an event happening can be
estimated experimentally, by repeating an experiment over
and over again. The probability is estimated using:
number of times event occurs
number of times experiment is repeated
This is referred to as the relative frequency.
You can increase the accuracy of the relative
frequency as an estimate of probability, by increasing
the number of times you repeat the experiment.
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Estimating probability
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Venn diagrams
Venn diagrams can be used to represent probabilities.
The outcomes that
satisfy event A can be
represented by a circle.
A
The outcomes that satisfy
event B can be represented
by another circle.
B
The circles can be overlapped to represent
outcomes that satisfy both events.
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Addition properties
Contents
Probability basics and notation
Estimating probability
Addition properties
Independent events
Conditional probability
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Addition properties
Two events A and B are called mutually exclusive if they
cannot occur at the same time.
For example, if a card is picked at random from a
standard pack of 52 cards, the events “the card is a club”
and “the card is a diamond” are mutually exclusive.
However the events “the card is a club” and “the
card is a queen” are not mutually exclusive.
If A and B are mutually exclusive,
then:
P(A B ) = P(A) + P(B )
A
B
In Venn diagrams
representing mutually
exclusive events, the circles
do not overlap.
This symbol means
‘union’ or ‘OR’
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Addition properties
This addition rule for finding P(A  B) is not true when
A and B are not mutually exclusive.
The more general rule for finding P(A  B) is:
P(A  B) = P(A) + P(B) – P(A  B)
This symbol means
‘intersect’ or ‘AND’
Venn diagrams can
help you to visualize
probability calculations.
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Addition properties
Example: A card is picked at random from a pack of cards.
Find the probability that it is either a club or a queen or both.
Card is a club = event C
Card is a queen = event Q
1
P(C ) =
4
4
1
P(Q ) =
=
52
13
P(C Q ) =
This represents the
other 3 queens.
This area
represents the
12 clubs that
are not queens.
1
52
This represents the
queen of clubs.
1
1
1
4
+
–
=
So, P(C Q ) =
4
13
52
13
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Addition properties
Example 2: If P(A′  B′) = 0.1, P(A) = 0.45 and P(B) = 0.75,
find P(A  B).
P(A′  B′) is the unshaded area in
the Venn Diagram.
A
B
We can deduce that:
0.1
P(A  B) = 1 – 0.1 = 0.9
Using the formula, P(A  B) = P(A) + P(B) – P(A  B),
we get:
0.9 = 0.45 + 0.75 – P(A  B)
0.9 = 1.2 – P(A  B)
So, P(A  B) = 0.3
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Addition properties
Examination-style question: There are two events, C and D.
P(C) = 2P(D) = 3P(C  D). Given that P(C  D) = 0.52, find:
a) P(C  D)
b) P(C  D′).
C
a) Let P(C  D) = x
D
x
P(C  D) = P(C) + P(D) – P(C  D)
So, 0.52 = 3x + 1.5x – x
Therefore x = P(C  D) = 0.52 ÷ 3.5
= 0.149 (3 s.f.)
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Addition properties
Question (continued):
C
D
b) P(C  D′) corresponds to the
unshaded area in this Venn diagram.
We see that:
P(C  D′) = P(C′  D′) + P(C)
= (1 – 0.52) + 3 × 0.149 …as P(C) = 3P(C  D)
= 0.48 + 0.447
= 0.927 (3 s.f.)
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Independent events
Contents
Probability basics and notation
Estimating probability
Addition properties
Independent events
Conditional probability
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Independent events
Two events are said to be independent if the occurrence of
one has no effect on the probability of the second occurring.
For example, if a coin and a die are both thrown, then
the events “the coin shows a head” and “the die shows
an odd number” are independent events.
If A and B are independent, then:
P(A  B) = P(A) × P(B)
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Independent events
Example: A and B are independent
events. P(A) = 0.7 and P(B) = 0.4.
a) Find P(A  B).
A
B
b) Find P(A′  B).
a) As A and B are independent,
P(A  B) = P(A) × P(B)
= 0.7 × 0.4 = 0.28
b) P(A′  B) is the shaded region in the Venn diagram.
So, P(A′  B) = P(B) – P(A  B)
= 0.4 – 0.28 = 0.12
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Independent events
Tree diagrams are sometimes a useful way of finding
probabilities that involve a succession of events.
Example: A bag contains 6 green counters and 4 blue
counters. A counter is chosen at random from the bag
and then replaced. This is repeated two more times.
Find the probability that the 3 counters chosen are
a) all green
b) not all the same colour.
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Independent events
0.6
0.4
0.6
0.4
0.4
0.6
B
G
0.4
0.6
B
G
0.4
B
0.6
G
0.4
B
B
G
B
0.4
G
G
G
0.6
0.6
B
a) P(GGG) = 0.6 × 0.6 × 0.6
= 0.216
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Independent events
0.6
0.4
0.6
0.4
0.4
0.6
B
G
0.4
0.6
B
G
0.4
B
0.6
G
0.4
B
B
G
B
0.4
G
G
G
0.6
0.6
B
b) To find the probability that there will be at least one
of each colour, we can find 1 – P(GGG) – P(BBB)
P(BBB) = 0.4 × 0.4 × 0.4 = 0.064.
Therefore, the answer is 1 – 0.216 – 0.064 = 0.72
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Conditional probability
Contents
Probability basics and notation
Estimating probability
Addition properties
Independent events
Conditional probability
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Conditional probability
However, the probability of event B happening might
depend on whether A has happened or not.
For example, if blue and green counters are pulled from a
bag twice and not replaced, then the probability of pulling
out a green counter on the second try will depend on what
colour was pulled out on the first try.
The probability that event A will happen, given that
event B has happened, is written
P(A | B)
This is a conditional probability.
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Conditional probability
To find the probability of events A and B both happening we
use:
P(A  B) = P(A) × P(B | A)
P(B | A)
B
P(A  B) = P(A) × P(B | A)
P(B′ | A)
B′
P(A  B′) = P(A) × P(B′ | A)
P(B | A′ )
B
P(A′  B) = P(A′) × P(B | A′)
P(B′ | A′ )
B′
P(A′  B′) = P(A′) × P(B′ | A′)
A
P(A)
P(A′)
A′
This formula can be re-arranged to give:
P(A B )
P(B | A) 
P(A)
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Conditional probability
Example: A bag contains 8 dark chocolates and 4 milk
chocolates. One chocolate is taken out and eaten. A second
chocolate is then taken. Find the probability that:
a) two milk chocolates are taken.
b) the two chocolates are of different types.
2
3
1
3
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7
11
D
4
11
M
8
11
D
3
11
M
D
M
1 3
1


a) P(M  M) =
3 11 11
2 4
8
b) P(D  M) =  
3 11 33
1 8
8
P(M  D) =  
3 11 33
8
8 16


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Conditional probability
Examination-style question: A man has 2 shirts
(one white and one blue) and 2 ties (red and silver).
If he wears the white shirt, he chooses the red tie
with probability 0.4.
If he wears the blue shirt, he chooses the red tie with
probability 0.75.
The probability that he wears the white shirt is 0.7.
a) Find the probability that he wears the red tie.
b) Given that he is wearing a red tie, find the
probability that he picked the blue shirt.
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Conditional probability
Tie
Shirt
0.4
R
0.6
S
0.75
R
0.25
S
W
0.7
0.3
B
a) P(red tie) = P(W  R) + P(B  R)
= (0.7 × 0.4) + (0.3 × 0.75)
= 0.28 + 0.225 = 0.505
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Conditional probability
b) Recall the formula for conditional probability:
P(A B )
P(B | A) 
P(A)
P(B R ) 0.225

So, P(B | R ) 
P(R )
0.505
= 0.446 (3 s.f.)
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Conditional probability
The example below demonstrates an important application of
probability in the field of medicine.
Example: A disease affects 1 in 500 people.
A diagnostic test for the disease records a positive
result 99% of the time when the disease is present
(this is called the sensitivity of the test).
The test records a negative result 95% of the time
when the disease in not present.
The test results are always either positive or negative.
Find the probability that a person has the disease,
given that the test result is positive.
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Conditional probability
Disease
Test
0.99
+ve
0.01
–ve
0.05
+ve
0.95
–ve
D
0.002
0.998
D′
P(D  +ve) = 0.002 × 0.99
= 0.00198
P(D′  +ve) = 0.998 × 0.05
= 0.0499
Therefore, P(+ve) = 0.05188
P(D +ve) 0.00198

 0.0382 (3 s.f.)
So, P(D |+ve) 
P(+ve)
0.05188
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