Fundamentals of Probability and Statistics for Engineers

where B 1 ,B 2 ,andB 3 are mutually exclusive, each occurring with probability^14.
It is easy to calculate the following:

We see that Equation (2.17) is satisfied for every j and k in this case, but
Equation (2.18) is not. In other words, events A 1 ,A 2 ,andA 3 are pairwise
independent but they are not mutually independent.
In general, therefore, we have Definition 2.2 for mutual independen ce of
nevents.

D ef inition 2. 2. Events A 1 ,A 2 ,...,An are mutually independent if and only if,
with k 1 ,k 2 ,...,km being any set of integers such that 1 k 1 <k 2 ...< km n
and m 2, 3,... , n,

The total number of equations defined by Equation (2.19) is 2n n1.

Example 2.8.Problem: a system consisting of five components is in working
order only when each component is functioning (‘good’). Let Si,i 1,...,5,be
the event that the ith component is good and assume P(Si)pi.Whatisthe
probability q that the system fails?
Answer: assuming that the five components per fo rm in an independent
manner, it is easier to determine q through finding the probability of system
success p. We have from the statement of the problem

Equation (2.19) thus gives, due to mutual independen ce of S 1 ,S 2 ,...,S 5 ,

H ence,

Basic Probability Concepts 19

P…A 1 †ˆP…B 1 [B 2 †ˆP…B 1 †‡P…B 2 †ˆ

1

2

;

P…A 2 †ˆP…A 3 †ˆ

1

2

;

P…A 1 A 2 †ˆP‰…B 1 [B 2 †\…B 1 [B 3 †ŠˆP…B 1 †ˆ

1

4

;

P…A 1 A 3 †ˆP…A 2 A 3 †ˆ

1

4

;

P…A 1 A 2 A 3 †ˆP‰…B 1 [B 2 †\…B 1 [B 3 †\…B 2 [B 3 †ŠˆP…;†ˆ 0 :

 

ˆ

P…Ak 1 Ak 2 ...Akm†ˆP…Ak 1 †P…Ak 2 †...P…Akm†: … 2 : 19 †



ˆ

ˆ

pˆP…S 1 S 2 S 3 S 4 S 5 †:

pˆP…S 1 †P…S 2 †...P…S 5 †ˆp 1 p 2 p 3 p 4 p 5 : … 2 : 20 †

qˆ 1 pˆ 1 p 1 p 2 p 3 p 4 p 5 : … 2 : 21 †