34 2. FUNDAMENTAL RELIABILITY MATHEMATICS
Solution:
In this experiment, we will have a total 6 6 D 36 possible sample points. e event B
1
D
f
4
g
contains three possible outcomes:
.
1 C 3
/
;
.
2 C 2
/
, and .3 C 1/, so the probability of P
.
f
4
g
/
will be
P
.
f
4
g
/
D
3
36
:
In the same approach, the event B
2
D
f
7
g
will contain six possible outcomes:
.
1 C 6
/
;
.
2 C 5
/
;
.
3 C 4
/
,
.
4 C 3
/
;
.
5 C 2
/
, and
.
6 C 1
/
. e probability of P
.
f
7
g
/
will
be:
P
.
f
7
g
/
D
6
36
:
erefore, according to Equation (2.16), we have:
P
.
A
/
D P
.
f
4; 7
g
/
D P
.
f
4
g
/
C P
.
f
7
g
/
D
3
36
C
12
36
D
5
12
:
2.5.3 PROBABILITY OF UNION AND INTERSECTION OF TWO
EVENTS
If events E
1
and E
2
are two events (sets) of the same experiment, the probability of the union
of two events is:
P
.
E
1
[ E
2
/
D P
.
E
1
/
C P
.
E
2
/
P
.
E
1
E
2
/
: (2.17)
Figure 2.5 clearly shows that the doubled crossed area is the intersection of the event E
1
and
E
2
. In the probability of P
.
E
1
/
C P
.
E
2
/
, the doubled crossed area is counted twice.
E
1
E
2
Figure 2.5: e Venn diagram of the union of the event E
1
and E
2
.
We can rearrange Equation (2.17) to form the formula to calculate the probability of the
intersection of two events E
1
and E
2
as,
P
.
E
1
E
2
/
D P
.
E
1
/
C P
.
E
2
/
P
.
E
1
[ E
2
/
: (2.18)
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