This preview shows page 1. Sign up to view the full content.
Unformatted text preview: CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 Assignment 4 Solutions
1. Problem 3.38 in Ross
a. 2 out of 4 system
P(2 out of 4 system works)=P(>=2 work)=1P(only 1 works)P(none work) Can also do:
P(2 out of 4 system works)= P(only 2 work)+P(only 3 work)+P(all 4 work)
The point is that you always need to consider all the components. When you are saying only two
of them work, then it means 2 work and the other two do not. One possible outcome is 1&2
work but 3&4 do not work. If I denote it as W1W2W3cW4c, then the probability is
P(W1W2W3cW4c)= P(W1) P(W2) P(W3c) P(W4c) (By independence)=P1 P2 (1P3)(1P4) b. 3 out of 5 system
P(3 out of 5 system works)=P(>=3 components work)
=1P(none work)P(only 1 works) P(only 2 work) Can also do:
P(3 out of 5 system works)= P(only 3 work)+P(only 4 work)+P(all 5 work)
2. Problem 3.41 in Ross
C1=component 1 works
Use conditional probability: if C1 works then you know that system works CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 3. Problem 3.47 in Ross
For parts a and b find union of A and B, P(A or B)
a. Mutually exclusive:
b. Independent:
Where
For parts c and d find intersection of A, B, and C: P(A&B&C)
c. Independent:
d. Mutually exclusive:
4. a.
b.
5. Given are:
D: detect
DA: detect by A
DB: detect by B
G: gross polluter
System A:
,
,
System B:
,
,
,
a. Find P(GDc) for A only. Use Bayes’ Theorem b. Find P(GD) for B only. Use Bayes’ Theorem c. The question is saying that DAG and DBG are independent but NOT DA and DB.
i. Find CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 ii. Find d. Grossly polluting if either sensor indicates this. i. Given
1. Sensor A only Since 2. Sensor B only 3. Both sensors ii. False alarm rate – find , we can get CE 93 – Engineering Data Analysis
6. Prof. Joan Walker, Spring 2011 Bay Bridge repairs:
Total
days (T) days for repair A + Probabilities days for repair B (TA+TB) 13 6+7 14 6+8 7+7 15 6+9 7+8 8+7 7+9 8+8 16
17 Total probability 0.06 0.06 0.06 0.15 0.08 0.15 0.09 0.32 0.2 0.09 0.29 0.12 0.12 8+9 PMF of total duration T to finish both repairs 0.21 ...
View Full
Document
 Spring '08
 Staff

Click to edit the document details