2.5. SOME BASIC OPERATIONS OF PROBABILITY 39
From Equation (2.21), we have
P .B
j
A/ D
P
.
B A
/
P
.
A
/
D
0:70
0:75
D 0:933:
Example 2.20
In an experiment of rolling a dice twice, calculate the probability of showing number 4 in the
second rolling when the number 5 has been shown in the first rolling.
Solution:
Let us use event A to represent showing number 4 in the first rolling and event B showing num-
ber 5 in the second rolling. Since event A and event B, in this case, are statistically independent,
so we have
P .B
j
A/ D P
.
B
/
D 1=6:
We can also use Equation (2.22) to run the calculation. In the first rolling, we can have P
.
A
/
D
1
6
. In rolling a dice twice, we can have a total 36 possible outcomes and showing number 5 and
then 4 is one outcome, so we have P
.
A B
/
D
1
36
. erefore, we have:
P .B
j
A/ D
P
.
B A
/
P
.
A
/
D
1=36
1=6
D 1=6:
2.5.7 TOTAL PROBABILITY THEOREM
If B
1
, B
2
,…,B
n
are a set of mutually exclusive, exhaustive collective event, that is,
B
j
B
i
D ;I j ¤ i; i; j D 1; 2; : : : ; n
D B
1
[ B
2
[ : : : [ B
n
:
e probability of another event A can always be expanded as
P
.
A
/
D P
.
A B
1
/
C P
.
A B
2
/
C : : : C P
.
A B
n
/
D
n
X
iD1
P
.
A B
i
/
D
n
X
iD1
P
.
A
j
B
i
/
P
.
B
i
/
: (2.24)
40 2. FUNDAMENTAL RELIABILITY MATHEMATICS
Example 2.21
ere was a total of 50 students in one engineering class with 40 male students and 10 female
students. In their final grades, 12.5% of male students had a grade “A” and 10.0% of female had
a grade “A”. Calculate the probability of students with a grade “A”.
Solution:
Let us use event E to represent a student with a grade “A”, and the events B
1
and B
2
to
represent a male and a female student, respectively.
According to the given information, we have:
P
.
B
1
/
D
40
50
D 0:8I P
.
E
j
B
1
/
D 0:125I P
.
B
2
/
D
10
50
D 0:2I P
.
E
j
B
2
/
D 0:10:
According to Equation (2.24), we have:
P
.
E
/
D P
.
E
j
B
1
/
P
.
B
1
/
C P
.
E
j
B
2
/
P
.
B
2
/
D 0:8 0:125 C 0:2 0:1 D 0:12 D 12%:
Example 2.22
A company purchases one type of ball bearing from three different suppliers: 45% from supplier
B
1
, 35% from supplier B
2
, and 20% from supplier B
3
. According to the information provided
by the suppliers, the probability of qualified bearings from the suppliers B
1
, B
2
, and B
3
are
92%, 95%, and 98%, respectively. Calculate the probability of a qualified bearing purchased by
the company.
Solution:
Let event Q to represent a qualified bearing. From the given information, we have
P
.
B
1
/
D 0:45I P
.
Q
j
B
1
/
D 0:92
P
.
B
2
/
D 0:35I P
.
G
j
B
2
/
D 0:95
P
.
B
3
/
D 0:20I P
.
G
j
B
3
/
D 0:98:
According to Equation (2.24), the probability of a qualified bearing in the company will be:
P
.
G
/
D P
.
G
j
B
1
/
P
.
B
1
/
C P
.
G
j
B
2
/
P
.
B
2
/
C P
.
G
j
B
3
/
P
.
B
3
/
D 0:45 0:92 C 0:35 0:95 C 0:20 0:98 D 0:9425 D 94:25%:
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