5 Solved Problems on Engineering Reliability - Assignment 2 | MEM 361, Assignments of Mechanical Engineering

Material Type: Assignment; Class: Engineering Reliability; Subject: Mechanical Engr & Mechanics; University: Drexel University; Term: Unknown 1989;

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MEM 361 Engineering Reliability
Problem Set 2
1) Suppose that P{X}=0.32, P{Y}=0.44 and P{XY}=0.58. Answer the following questions
with proof.
(a) Are X and Y mutually exclusive?
(b) Are X and Y independent events?
(c) What is the value of P{X|Y}?
(d) What is the value of P{Y|X}?
Solution: a) ,
X
Y are mutually exclusive only if
(
)
0PX Y
=. Now
()
(
)
(
)
(
)
()()()()
0.32 0.44 0.58 0.18
PX Y PX PY PX Y
PX Y PX PY PX Y
∪= + ∩⇒
∩= + ∪= + =
So NOT mutually exclusive
b) For independent events
(
)
(
)
(
)
PX Y PXPY∩= , but this is not true, so NOT
independent
c)
()
()
(
)
0.18 / 0.44 0 .409PXY PX Y PY=∩ = =
d)
()
()
(
)
0.18 / 0.32 0.5625PYX P X Y PX=∩ = =
2) In a quality test of 126 computer chips from two different suppliers, the following are the test
results:
pass test fail test
Supplier -1 80 4
Supplier -2 40 2
Let A denote the event that a chip is from supplier-1 and B the event that a chip passes the quality
test. Answer the following questions.
(a) Are A and B independent events?
(b) Are Ac and B independent events?
(c) What is the meaning of P{AB}?
(d) What is the value of P{AB}?
Solution:
a)
() ()
(
)
()
()() ()
()
()() ()
84 / 126; 120 /126; 80 / 126
80 /120 2 / 3
80 / 84 20 / 21
PA PB PA B
PAB PA B PB PA
PBA PA B PA PB
==∩=
=∩ = ==
=∩ = = =
So ,
A
B ARE independent
b) ,
c
A
B ARE independent
c) The probability that A, B or both occur
d)
()()()()
84 120 80 124
126 126 126 126
PA B PA PB PA B∪= + ∩= + =
3) Use the Venn diagram and show the following equalities:
pf3
pf4

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MEM 361 Engineering Reliability

Problem Set 2

1) Suppose that P{X}=0.32, P{Y}=0.44 and P{X∪ Y}=0.58. Answer the following questions

with proof.

(a) Are X and Y mutually exclusive?

(b) Are X and Y independent events?

(c) What is the value of P{X|Y}?

(d) What is the value of P{Y|X}?

Solution:

a) X , Y are mutually exclusive only if P (^) ( XY ) = 0. Now ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.32 0.44 0.58 0.

P X Y P X P Y P X Y

P X Y P X P Y P X Y

So NOT mutually exclusive

b) For independent events P (^) ( XY (^) ) = P (^) ( X (^) ) P Y ( ), but this is not true, so NOT

independent

c) P (^) ( X Y (^) ) = P ( XY ) P Y ( ) = 0.18 / 0.44 =0. d) P Y X ( (^) ) = P ( XY ) P ( X ) = 0.18 / 0.32 =0.

2) In a quality test of 126 computer chips from two different suppliers, the following are the test

results:

pass test fail test

Supplier -1 80 4

Supplier -2 40 2

Let A denote the event that a chip is from supplier-1 and B the event that a chip passes the quality

test. Answer the following questions.

(a) Are A and B independent events?

(b) Are Ac^ and B independent events?

(c) What is the meaning of P{A∪ B}?

(d) What is the value of P{A∪ B}?

Solution:

a)

( ) ( ) ( ) ( ) (^ )^ (^ )^ (^ ) ( ) ( ) ( ) ( )

P A P B P A B

P A B P A B P B P A

P B A P A B P A P B

So A , B ARE independent

b) Ac , B ARE independent

c) The probability that A , B or both occur

d) P (^) ( AB ) (^) = P (^) ( A ) (^) + P B ( ) (^) − P (^) ( AB ) = 12684 + 120126 − 12680 =^124126

3) Use the Venn diagram and show the following equalities:

if and are independent events

c c c c

P Y P Y X P Y X

P Y P Y X P X P Y X P X

P Y P Y P X P Y P X X Y

Solution:

The first equality follows directly from the diagrams below.

S S S

P ( X ∩ Y ) P^ ( X^ c ∩^ Y ) P ( Y )

The second applies the definition of conditional probability.

If the events X and Y are independent, then, by definition,

P Y ( | X ) = P Y ( P^ (^ ∩ X ) X )= P Y ( )

So for independent X and Y the first equality becomes

P Y ( ∩ X c ) = P Y ( ) − P Y ( ) P ( X ) = P Y ( ) ( 1 − P ( X ))

Thus,

( |^ ) (^ ( ))^ (^1 )^ (^1 ( ()^ )) ( )

c^ c c

P Y X P Y^ X^ P Y^ P^ X P Y

P X P^ X

= ∩^ = − =

Therefore X c and Y are independent, in which case the second equality becomes the third.

4) An electric motor is used to power a cooling fan; an auxiliary battery is used in the event of a

main power outage. Experiment indicates that the chance of a main power outage is 0.6%. And,

when the main power is on, the chance that the motor itself fails is pm= 0.25x10-3; when the

auxiliary battery is on, the chance of the motor failure is pb = 0.75x10-3. Determine the probability

that the cooling fan fails to function.

Solution:

Let

event of a main power failure event of fan failing to operate

X

Y

The sample space divides into the two disjoint sets X and its complement X c so.

Thus, the expansion rule applies:

YXX^ c

P Y ( ) = P Y ( | X ) P ( X ) + P Y ( | X c^ ) P ( Xc )

Now,