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Probability Homework Solution: Lifetime of Chips as a Sequence of Bernoulli Trials, Exercises of Probability and Statistics

A solution to a probability homework problem involving the lifetime of chips, represented as a sequence of bernoulli trials. Calculations and matrix notation to determine the probability of a chip's lifetime exceeding a certain threshold and the expected number of trials required. It also discusses the concept of eigenvalues and eigenvectors in relation to the problem.

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

anandini
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Probability Homework Solution #4
2.71
2100 3
0
100 99 98 2
4
[3 or more errors] 1 [2 or fewer errors]
100
1(1)10
100 99
1 (1 ) 100(1 ) (1 )
2
1 099985 1.5 10
kk
k
PP
pp p
k
p
pp pp
−−
=
=−
⎛⎞
=− =
⎜⎟
⎝⎠
×
⎧⎫
=− + +
⎨⎬
⎩⎭
=− = ×
2.75
For an individual chip, the problem that the lifetime exceeds 1
α
seconds is:
11
1
lifetimePeep
αα
α
−⋅
⎡⎤
>= =
⎢⎥
⎣⎦
If we consider the testing of whether each chip lifetime exceeds 1
α
as a sequence of
Bennoulli trials, then
[]
10 1110
5
10
5 ( ) (1 ) 0.289
kk
k
Pk e e
k
−−
=
⎛⎞
≥= =
⎜⎟
⎝⎠
2.82
(a)
01
11
(1) , (1)
22
pp==
(b)
001
101
21
(1) () ()
36
15
(1) () ()
36
p
npnpn
p
npnpn
+= +
+= +
In matrix notation, we have
[][]
01 01
21
33
(1),(1) (),()
15
66
pn pn pnpn
++=
, or (1) ()
p
npn
+
=Ρ
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Download Probability Homework Solution: Lifetime of Chips as a Sequence of Bernoulli Trials and more Exercises Probability and Statistics in PDF only on Docsity!

1

Probability Homework Solution

2

100 3

0

100 99 98 2

4

[3 or more errors] 1 [2 or fewer errors]

k k

k

P P

p p p

k

p p p p p

− −

=

×

= − = ×

For an individual chip, the problem that the lifetime exceeds

seconds is:

1

1

P lifetime e e p

α

α

− ⋅

If we consider the testing of whether each chip lifetime exceeds

as a sequence of

Bennoulli trials, then

[ ]

10

1 1 10

5

k k

k

P k e e

k

− − −

=

(a)

0 1

p = p =

(b)

0 0 1

1 0 1

p n p n p n

p n p n p n

In matrix notation, we have

[ ] [ ]

0 1 0 1

p n p n p n p n

, or p ( n + 1) = p n ( ) Ρ

2

(c)

p

p

p

p

p

(d)

2

p

p p

p p p p

P

P PP P

in general ( ) (0)

n

p n = p P. To find

n

P , we note that if P has eigenvalues

1

2

and eigenecotrs

1

e ,

2

e , then

1 1

1 2

2

, where [ , ]

e e

Λ

P E E E

, and

1 1 1

1 -1 1

1

( )( ) ( ) times

n

n

n

− − −

− −

P E E E E E E

E E E E E E

E E