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PROBABILITY
1. The Terminology of Probability
Suppose there are 100 students in a school sixth form, 50 studying mathematics, 29 studying
biology, and 13 studying both. We can summarise this information in a Venn diagram.
Biology
Mathematics
37
13
16
34
Notice how the 13 who study both subjects are represented, and also the 34 who study neither
subject.
Suppose we pick a student at random. We introduce the following symbols...
symbol
meaning
example
P( M ) : the probability that the student
′
not
does not study mathematics.
P( M  B) : the probability that the student
and

(intersection) studies both mathematics and biology.
P( M  B) : the probability that the student
or

(union)
studies mathematics or biology or both.
So we have, for example, the following probabilities...
71
13
37
P( B) 
P( M  B ) 
P( M  B) 
100
100
100
66
84
16
P( M  B ) 
P( M  B) 
P( M   B) 
100
100
100
Note that P( B)  1  P( B).
2. Exhaustive Events
Two (or more) events A and B are exhaustive if, together, they exhaust all the possibilities. That
is, P( A  B)  1 . An example of such a pair of events is
• A = a new-born baby is male
• B = a new-born baby is female
For exhaustive events, the region outside the circles on a Venn diagram is empty.
3. Mutually Exclusive Events
Two (or more) events A and B are mutually exclusive if they cannot happen together. That is,
P( A  B)  0 . An example of such a pair of events is
• A = Manchester United win the match
• B = Manchester United lose the match
For mutually exclusive events, the intersection on a Venn diagram is empty.
4. The Addition Rule
From our Venn diagram, we can see that the probability that a random student studies either
mathematics or biology (or both) is given by
P( M  B)  P( M )  P( B)  P( M  B)
66
50
29
13



100
100 100
100
We subtract P( M  B) at the end, since the probability of the student studying mathematics and
the probability of the student studying biology each include the probability of the student
studying both mathematics and biology. We only need this probability once, so the extra case is
subtracted. Diagrammatically,
+
=
–
In general, we have the addition rule for events A and B...
P( A  B)  P( A)  P( B)  P( A  B)
Example 1 : In a group of 20 adults, 4 out of the 7 women and 2 out of the 13 men wear glasses.
What is the probability that a person chosen at random from the group is a woman
or someone who wears glasses?
P(W  G )  P(W )  P(G)  P(W  G)
7
6
4

 
20 20 20
9

20
Example 2 : In a street containing 20 houses, 3 households do not own a television set, 12
households have a black and white set, and 7 households have a colour and a black
and white set. Find the probability that a household chosen at random owns a colour
television set.
P(C  B)  P(C )  P( B)  P(C  B)
17
12 7
 P(C )  
20
20 20
12
P(C ) 
20
Example 3 : Given that P( A)  23 , P( B)  12 and P( A  B)  121 , find P( A  B) .
P( A  B)  P( A)  P( B)  P( A  B)
 13  12  121
 129
Some people prefer the visual approach of a Venn diagram rather than the addition rule.
Example 4 : A card is selected at random from a regular set of 52 playing cards. Let Q be the
event that the card is a queen and D be the event that the card is a diamond.
Find
a) P(Q  D )
b) P(Q  D )
c) P(Q  D)
d) P(Q  D) .
We first draw the Venn Diagram for the 52 cards.
Q
D
3
1
12
36
P(Q  D) 
1
52
P(Q  D) 
16
52
P(Q  D) 
49
52
P(Q  D) 
3
52
p73 Ex 5A
5. The Multiplication Rule
To find the probability of two (or more) events happening together, we multiply the individual
probabilities of the events together.
P( A  B)  P( A)  P( B)
This equation is not quite correct, but will do for now! We will make it watertight in section 7.
Example 1 : Find the probability of throwing a six with a dice, and a coin landing heads up.
P(6  H )  P(6)  P( H )
1 1
 
6 2
1

12
Example 2 : I have 3 black and 2 red socks in a drawer. I take two out at random. Find the
probability I pick out two black socks.
P( B1  B2 )  P( B1 )  P( B2 )
3 2
 
5 4
6

20
Note the probability of getting a black sock the second time round is affected by what happened
on the first pick. This is known as a conditional probability : more on these later.
Example 3 : Find the probability of winning the National Lottery jackpot.
P( jackpot)  P(1st)  P(2nd)  P(3rd)  P(4th)  P(5th)  P(6th)
6 5 4 3 2 1
     
49 48 47 46 45 44
1

13983816
Example 4 : An A-level maths class contains 5 boys and 7 girls. As an example to the rest, three
students are taken out and flogged. Find the probability that at least one boy is
flogged.
This examples illustrates an important shortcut. Rather than add the probabilities of one boy, two
boys and three boys being flogged, we do this...
P(at least one boy)  1  P(no boys)
 1  P(all girls)
7 6 5
 
12 11 10
7
 1
44
37

44
 1
6. Tree Diagrams
These make the multiplication rule easier to visualise. They are particularly useful when there is
more than one route to a successful outcome.
Example 1 : Bag A contains 3 red and 2 blue counters. Bag B contains 1 red and 3 blues. A
counter is drawn from each bag find the probability that they are different colours.
3
5
2
5
1
4
red RR
3
20
3
4
blue RB
9
20
1
4
red BR
2
20
blue BB
6
20
red
blue
3
4
9
2

20 20
11

20
P(different colours) 
Example 2 : Mrs. Smith catches a bus to the station. The bus is late with probability 0·4. If the
bus is late she catches her train with probability 0·2. If the bus is on time she catches
her train with probability 0·5. Find the probability that she catches her train.
We define the events
• L : the bus is late
• T : Mrs. Smith catches the train
0.2
T
LT
0.8
T'
LT ' 0.32
0.5
T
L'T
0.5
T ' L' T ' 0.3
0.08
L
0.4
0.6
0.3
P(T )  P( L  T )  P( L  T )
 0.08  0.3
 0.38
L'
Example 3 : I have three pound coins and four ten pence pieces in my pocket. I pull out two coins
at random. Find the probability both coins are the same.
3
7
2
6
£1
6
42
4
6
10p
12
42
3
6
£1
12
42
10p
12
42
£1
4
7
10p
3
6
P(same)  P(£1  £1)  P(10p  10p)
6 12


42 42
18

42
We can make things easier by only drawing the branches that are necessary.
Example 4 : I have two red and three black socks in a drawer. I pull out two socks at random.
Find the probability they are a matching pair.
2
5
red
3
5
black
1
4
3
4
red
2
20
black
9
20
2
6

20 20
8

20
P(matching pair) 
Example 5 : Sally has nine pens in her drawer : four red, three blue and two green. She takes
three out at random. Find the probability they are all different colours.
4
RBG
32  1
9
8
7
21
RGB
4
9
   211
BRG
3
9
 84  72  211
BGR
3
9
 82  74  211
GRB
2
9
 84  73  211
GBR
2
9
 83  74  211
2
8
3
7
6
21
2

7
P(all different) 
7. Conditional Probability
Consider again the Venn diagram from section 1.
Biology
Mathematics
37
13
16
34
Suppose that we pick a mathematics student at random, and try to find the probability that this
student also studies biology. This probability is written as P( B | M ) and is read as ‘the
probability of B given M’. This is known as a conditional probability - the probability that a
student studies biology, given the condition that the student already studies mathematics.
In this case we have
P( B | M ) 
13
50
since 13 of the biology students also study mathematics.
Notice that
P( B | M ) 
13
50

 13 


 100 
 50 


 100 

P( B  M )
P( M )
..which makes sense if you think about it long enough.
Rearranging, we have
P( B  M )  P( B | M )  P( M )
In general, we have the multiplication rule for two events A and B...
P( B  A)  P( B | A)  P( A)
...and this is the watertight version of the multiplication rule we referred to in section 5!
We already met this idea, albeit less formally, in example 3 of the previous section. The
probabilities on the seconds branches are all conditional probabilities - they differ depending on
what the first outcome is. So, for example, the 64 is actually P(10p | £1) . And in example 4, the 14
is actually P(red | red) .
The general conditional probability formula is
P( B | A) 
P( B  A)
P( A)
Example 1 : I post a letter, choosing a stamp at random. The probability I choose a first-class
stamp is 0.7. If I choose a first-class stamp, the probability the letter is delivered the
next day is 0.9, otherwise the probability is 0.5. Find the probability that I put a firstclass stamp on the letter, given that it arrives the next day.
We define the events
• F : the stamp is first-class
• N : the letter arrives the next day
FN
0.9
N
0.1
N ' FN ' 0.07
0.5
N
0.5
N ' F ' N ' 0.15
0.63
F
0.7
0.3
F 'N 0.15
F'
We require P( F | N ) .
P( F  N )
P( N )
0.63

0.63  0.15
63

78
P( F | N ) 
Example 2 : The probability a person has a rare blood disease is 1%. A new test for the disease is
introduced. The probability it yields a positive result for an infected person is 95%.
The probability it yields a positive result for an uninfected person is 2%.
Find the probability that if the result is positive, the person is infected. Comment on
the effectiveness of the test.
We define the events
• D : the person has the disease
• P : the test is positive
DP
0.95
P
0.05
P ' DP '
0.0005
0.02
P
D 'P
0.0198
0.98
P ' D ' P '
0.0095
D
0.01
0.99
D'
0.9702
We require P( D | P)
P( D  P)
P( P)
0.0095

0.0095  0.0198
95

293
 32%
P( D | P) 
A positive result is not a reliable indicator as to whether the person has the disease. There is a
32% chance they have the disease, and a 68% chance they don’t!
Example 3 : An apprentice welder takes two examinations - a practical and a written. To pass, an
apprentice must pass both exams. One year there are 60 apprentices. 35 pass the
practical exam and 22 pass the written exam. One-third of the welders pass neither
exam. Find the probability an apprentice
a) passes both exams,
b) passes the written exam, given they have passed the practical,
c) fails only the written exam, given that they have failed overall.
a) A Venn diagram is the best method here. Now 13  60  20 apprentices fail both exams, so 40
pass one exam or both. Using the addition rule,
P( P  W )  P( P)  P(W )  P( P  W )
40 35 22

  P( P  W )
60 60 60
17
P( P  W ) 
60
We can now fill in our Venn diagram.
Written
Practical
18
17
5
20
b)
c)
P(W  P)
P( P)
17

35
P(W   P  fail)
P(W   P | fail) 
P(fail)
18

43
P(W | P) 
Example 4 : There are two bags each containing the numbers from 1 to 10. In bag A, the prime
numbers are blue and the rest red. In bag B the square numbers are blue and the rest
red.
A bag is chosen at random, and a number removed. The number is not replaced.
Once again, a bag is chosen at random, and a number removed.
Find the probability that
a) both numbers are blue,
b) given that both numbers are blue, bag B was used on both occasions.
Drawing only those branches of the tree diagram that we need,
A
1
2
1
2
a)
B
4
10
3
10
1
2
blue
1
2
B
1
2
A
blue
12
12
12
6



360 400 400 360
11

100
P(both blue) 
A
1
2
B
3
9
3
10
4
10
2
9
blue
12
360
blue
12
400
blue
12
400
blue
6
360
b)
P(both bag B | both blue) 
P(both bag B  both blue)
P(both blue)

 6 


 360 
 11 


 100 

5
33
Example 5 : There are two A-level maths classes in Year 12. Group X contains eleven girls and
seven boys. Group Y contains six girls and eight boys.
Two students are chosen at random. Find the probability that
a) they are of the same gender,
b) they are from the same group,
c) they are of the same gender and from the same group,
d) they are of the same gender, given that they are from the same group,
e) they are from the same group, given that they are of the same gender.
Let B1 represent the event that the first choice is a boy, and X1 represent the event the first
choice is from group X. Also, let BX1 represent the event that the first choice is a boy from
group X.
a)
15 14 17 16
P( B1  B2 )  P(G1  G2 )    
32 31 32 31
482

992
b)
18 17 14 13
P( X 1  X 2 )  P(Y1  Y2 )    
32 31 32 31
488

992
c)
7 6
8 7 11 10 6 5
P( BX 1  BX 2 )  P( BY1  BY2 )  P(GX 1  GX 2 )  P(GY1  GY2 ) 


32

31
238
992
d)
P(same gender | same group) 

P(same gender  same group)
P(same group)
 238 


 992 
 488 


 992 
238
488
P(same group  same gender)
P(same group | same gender) 
P(same gender)

e)

 238 


 992 
 482 


 992 

238
482

32

31

32

31

32
31
Example 6 : For a married couple, the probability that the husband has passed his driving test is
7
, and the probability that the wife has passed is 12 . The probability that the
10
husband has passed given that the wife has passed is
14
15
. Find the probability that for
a couple chosen at random the driving test will have been passed by
a) both of them,
b) only one of them,
c) neither of them,
d) the wife, given that the husband has not passed.
e) If two married couples are chosen at random, find the probability that exactly one
of the husbands and exactly one of the wives will have passed the test.
a) Using the multiplication rule,
P( H  W )  P( H | W )  P(W )
14 1
 
15 2
14

30
At this stage it is convenient to draw a Venn diagram for 30 trials (the LCM of 2 and 15).
H
W
7
14
1
8
b)
P( H  W )  P( H   W ) 
c)
P( H   W ) 
d)
1
9
There are four possibilities (assuming the two couples are independent)
14 8 112
P( H1  W1 )  P( H 2  W2 )   
30 30 900
7 1
7
P( H1  W1)  P( H 2  W2 )   
30 30 900
1 7
7
P( H1  W1 )  P( H 2  W2 )   
30 30 900
8 14 112
P( H1  W1 )  P( H 2  W2 )   
15 15 900
Adding to find the total probability,
112
7
7 112 238




900 900 900 900 900
e)
7
1

30 30
8

30
P(W | H ) 
8
30
Example 7 : At a football match, 70% of male fans support United, and 60% of female fans
support City. The probability that a fan, selected at random, supports United is 0.67.
a) Find the probability that a fan, selected at random, is female.
b) Given that a fan supports City, find the probability that the fan is male.
Although this problem can be solved without it, a tree diagram makes it clearer. We introduce p
for the probability that a fan is male.
0.7
U
MU
0.7p
0.3
C
MC
0.3p
0.4
U
FU
0.4(1 – p)
0.6
C
FC
0.6(1 – p)
M
p
1 –p
F
Now
P(U )  0.67
0.7 p  0.4(1  p)  0.67
0.3 p  0.4  0.67
0.3 p  0.27
p  0.9
Therefore
a)
P( F )  0.1
b)
P( M | C ) 
P( M  C )
P(C )
0.9  0.3

0.9  0.3  0.1 0.6
0.27

0.33
27

33
9

11
8. Independent Events
If two events are independent, they have no effect on the probability of one another. For example,
the event ‘it rains’ and the event ‘the indoor bowls is postponed’ are independent. The events ‘it
rains’ and ‘the Wimbledon final is postponed’ are certainly not independent!
Suppose two events A and B are independent, and that event B has actually happened. Because A
and B are independent,this does not affect the probability of A happening. So we can say
P( A | B)  P( A) for independent events
This changes our multiplication rule from P( A  B)  P( A | B)  P( B) to
P( A  B)  P( A)  P( B) for independent events
and this is our simple version of the multiplication rule from section 5!
Consider once again the Venn diagram from section 1.
Biology
Mathematics
37
13
16
34
There are several ways to find whether the choice of biology and mathematics are independent.
50
• P(M )  100
 0.5 whereas P(M | B)  13
 0.448 , so students are less likely to choose
29
mathematics if they have already chosen biology.
29
• P( B)  100
 0.29 whereas P( B | M )  13
 0.26 , so students are less likely to choose
50
biology if they have already chosen mathematics.
13
29
50
• P( B  M )  100
 0.13 , whereas P( B)  P(M )  100
 100
 0.145 , which is not the same.
There is not a great deal of discrepancy in the above results, so we could say that the choice of
biology and mathematics are not heavily dependent.
Example 1 : In a certain group of 15 students, 4 have graphic calculators and 2 have a computer
at home. In addition, one student has both. Show that the events A : ‘the student has
a graphic calculator’ and B : ‘the student has a computer at home’ are independent.
1
15
5 3
P( A)  P( B)  
15 15
1

15
Since P( A  B)  P( A)  P( B) , the two events are independent.
P( A  B) 
Activity 1 : Use the two other methods to test for independence.
p85 Ex 5C
p90 Ex 5D
p92 Ex 5E
Topic Review : Probability