1
HW #2
Solutions (Fall 2011)
Assigned problems
:
2.24, 2.37, 2.48, 2.64, 2.65
2.24
(a)
Let A, B denote the event of the respective engineers spotting the error. Let E denote the event that
the error is spotted, P(E) = P(A
∪
B)
= P(A) + P(B) – P(AB)
= P(A) + P(B) – P(A)P(B)
= 0.8 + 0.9 – 0.8
×
0.9 =
0.98
(b)
“Spotted by A alone” implies that B failed to spot it, hence the required probability is
P(AB’E) = P(AB’
∩
E)/P(E)
= P(EAB’)P(AB’) / P(E)
= 1
×
P(A)P(B’) / P(E)
= 0.8
×
0.1 / 0.98
≅
0.082
(c)
With these 3 engineers checking it, the probability of not finding the error is
P(C
1
’C
2
’C
3
’) = P(C
1
’)P(C
2
’)P(C
3
’) = (1 – 0.75)
3
, hence
P(error spotted) = 1 –
(1 – 0.75)
3
≅
0.984,
which is higher than the 0.98 in (a), so the team of 3 is better.
(d)
The probability that the first error is detected (event D
1
) has been calculated in (a) to be 0.98.
However, since statistical independence is given, detection of the second error (event D
2
) still has
the same probability. Hence P(D
1
D
2
) = P(D
1
)P(D
2
) = 0.98
2
≅
0.960
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2.37
(a)
Let subscripts 1 and 2 denote “after first earthquake” and “after second earthquake”. Note that (i)
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 '11
 MTODOROVSKA
 Probability theory, mutually exclusive events, heavy damage

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