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1.4
The Axioms of Probability
Frequency definition of probability
Equally Likely Outcomes
Definition 1.9
A Probability Measure on a sample space S is a function
P which assigns a number P(A) to every event A in S in
such a way that the following three axioms are satisfied:
Axiom 1. P(A)≥0 for every event A.
Axiom 2. P(S) =1
Axiom 3. Countable Additivity i.e. if A1,A2,…is an
infinite sequence of mutually exclusive (disjoint) events
then

P(
i 1
[or P ( A1
A2

Ai )   P ( Ai )
i 1
)  P ( A1 )  P ( A2 )  ]
Example 1.10
A coin is tossed twice.
What is the probability that at least one head occurs?
Solution:
Example 1.11
A die is loaded in such a way that an even number is
twice as likely to occur as an odd number.
If E is the event that a umber less than 4 occurs on a
single toss of the die, find P(E).
Solution:
Example 1.12
In Example 1.11, let A be the event that an even number turns up and let B be the event that a number divisible by 3 occurs. Find P ( A B) and P ( A B)
Solution:
Properties of Probability
1) P(Ø)=0.
2) Finite Additivity
i.e. if A1 , A2 ,, An are mutually exclusive (disjoint) events
 n  n
then P   Ai    P ( Ai )
 i 1  i 1
3) P ( AC )  1  P ( A).
4) If B  A then P ( A  B )  P ( A)  P ( B ) and P ( B )  P ( A)
5) P(A)≤1 for all events A.
6) If A and B are arbitrary events then
P ( A  B )  P ( A)  P ( B )  P ( A  B )
and hence P ( A  B )  P ( A)  P ( B ).
Equally Likely Outcomes
Theorem
Let S be the sample space of an experiment.
If S has n elements which are all equally likely,
then for any event
A S
| A|
P ( A) 
n
where |A| denotes the number of elements in A.
Model One: Drawing ball
(1) Drawing without replacement
Problem 1 There are 4 white balls and 2 black balls
in a bag. Draw two balls consecutively out of the bag. Now
find the probability when the two balls are white.
Solution:
(2) Drawing with replacement
Problem 2 There are 4 red balls and 6 black balls in
a bag. Draw consecutively 3 balls with replacement. Find
Find the probability when the first two balls drawn are
black and the third ball drawn is red.
Solution:
Class Exercise
1o Phone number Given that a phone number is
made up of 7 digits, while the first digit can not be 0,
find the probability that the number 0 appears exactly
3 times.
( Solution : p  C 91  C 63 93 9  106 )
2o Dice 3 die are cast, find the probability that
the numbers you get after you throw the die add up to 4.
( Solution : p  3 63 )
Model Two: Ball placed in cylinder
(1) No restriction on the capacity of cylinder
Problem 1 Four balls are to be placed inside three
cylinders. Assuming that there is no restriction on the
number of balls each cylinder can contain, find the
probability that cylinders 1 and 2 each contains 2 balls.
Solution:
(2) Each cylinder can only have 1 ball
Problem 2 Four balls are to be placed into ten
cylinders. Knowing that each cylinder can only contain
one ball, find the probability that for the first 4 cylinders,
each one contains a ball.
Solution:
Class Exercise
1o Room arrangement There are three people Mr.
A, B and C who are going to be arranged into three
rooms. Find the probability that each room has just one
people in it.
2
( Solution : )
9
2o Birthday A class has a total of 20 students, who
are all born in the same year. Find the probability that
out of the 20 students, 10 were born on January 1st and
10 were born on December 31st.
 20  10 
( Solution : p     36520 )
 10  10 
Solution: The sample space for this experiment is
S  { HH , HT ,TH ,TT }
Therefore, we assign a probability of w to each sample
point. Then 4w=1,or w=1/4.
If A represents the event of at least one head occuring
then A  { HH , HT ,TH }
Thus
你认为哪个
结果出现的
可能性大？
1 1 1 3
P ( A)     .
4 4 4 4
Solution: The sample space is S={1,2,3,4,5,6}.
We assign a probability of w to each odd number and
a probability of 2w to each even number.
Since the sum of the probabilities must be 1, we have
9w=1 or w=1/9.
Hence probabilities of 1/9 and 2/9 are assigned to
each odd and even number, respectively. Therefore,
and
E  {1,2,3}
P ( E )  1 / 9  2 / 9  1 / 9  4 / 9.
Solution: For the events A={2,4,6} and B={3,6},
we have
A B  {2,3,4,6} and A B  {6}
By assigning a probability of 1/9 to each odd number
and 2/9 to each even number, we have
P( A B)  2 / 9
and
P( A
B)  2 / 9  1 / 9  2 / 9  2 / 9  7 / 9
Solution: Suppose A  {both balls are white},
The number of all events in the sample space is C 62
The number of basic events in A is C 42 ,So
P ( A)  C 42 / C 62
Solution: Suppose
A  { Black on the first 2 draws, red on the third draw }
Red on the third draw 4
First
draw
second
draw
Third
draw
6种
Black on
on the
the second
first draw
Black
draw 6
10
Total number of basic/simple events 10  10  10  103 ,
A contains 6  6  4  24 events, So
6 6 4
 0.144
P ( A) 
3
10
3
3
3
3
Solution: According to this type of placement, there
are 3  3  3  3  34 ways to place the 4 balls into 3 cylinders.

C42
2
C22
2
Therefore, the probability that cylinders 1 and
2 will each have 2 balls is:
2
2
2
4
p  C4  C2 3  .
27
Problem 2: Four balls are to be placed into ten
cylinders. Knowing that each cylinder can only contain
one ball, find the probability that for the first 4 cylinders,
each one contains a ball.
4  3  2 1
1
p 44

.
Solution: p  4 
p10 10  9  8  7 210