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ESI 6321
Optional Make-Up Exam
Solution Key
- See HW 4 solutions.
- See HW 4 solutions.
- The sample space values are 1, 2 , 3 , 4 , 5 , 6 , 7 , 8, and 9 after simple enumeration. Their corre-
sponding probabilities (obtained along the enumeration paths) are respectively:
(1) p^20 ; (2) p 0 p 2 + p^20 p 2 ; (3) p 0 p 3 + 2p 0 p^22 + p^30 p 3 ; (4) 2p 0 p 2 p 3 + p^32 + 3p^20 p 2 p 3 ;
(5) 2p^22 p 3 + 3p 0 p^22 p 3 + 3p^20 p^23 ; (6) p 2 p^23 + p^32 p 3 + 6p 0 p 2 p^23 ; (7) 3p 0 p^33 + 3p^22 p^23 ; (8) 3p 2 p^33 ; (9) p^43.
Obviously, this approach would be tedious if we wanted to generalize it. A more efficient
approach is based on the concept of Markov chain, which is not covered in this course.
- F (x) = 1 − P (X > x). Therefore
f (x) =
d
dx
F (x)
r∑− 1
k=
d
dx
e−λx(λx)k
k!
= λe−λx
∑r−^1
k=
(λx)k
k!
− e−λx
∑r−^1
k=
kλ(λx)k−^1
k!
= λe−λx^ + λe−λx
r∑− 1
k=
(λx)k
k!
− e−λx
r∑− 1
k=
kλ(λx)k−^1
k!
= λe
−λx − λe
−λx
r∑− 1
k=
[
k(λx)k−^1
k!
(λx)k
k!
]
= λe−λx
r∑− 1
k=
[
k(λx)k−^1
k!
(λx)k
k!
]}
= λe
−λx
[
(λx)r−^1
(r − 1)!
]}
λrxr−^1 e−λx
(r − 1)!
- (a) Define the following events:
E: product fails
EA: part A fails
EB : part B fails
EC : part C fails
We want to determine P {EB ∪ EC |E}.
Recall that the properties of unconditional probabilities apply also to conditional proba-
bilities. So we can write: P {EB ∪ EC |E} = 1−P
(EB ∪ EC )
′ |E
= 1−P {E B′ ∩ E′ C |E} =
1 − P {E B′ |E} P {E C′ |E}, where the last equality results from the independence assump-
tion. From the latter expression, we can further write:
P {EB ∪ EC |E} = 1 − (1 − P {EB |E}) (1 − P {EC |E})
Note that we could not use P {EB ∪ EC |E} = P {EB |E}+P {EC |E} at the outset as EB
and EC are not disjoint. Indeed, EB ∩ EC cannot be empty (and thus with probability
- since P {EB ∩ EC } = P {EB } P {EC } = 0. 0120
k=
k
- 01 k^ (1 − 0 .01)
20 −k
0 by
the independence assumption. (Strictly speaking, we should be checking that the events
EB and EC are conditionally (on E) not disjoint, but the argument is similar.)
The next step is to evaluate P {EB |E} and P {EC |E} through Bayes’ formula.
P {EB |E} =
P {E|EB } P {EB }
P {E}
1 × P {EB }
P {EA ∪ EB ∪ EC }
1 × P {EB }
1 − P
(EA ∪ EB ∪ EC )
1 × P {EB }
1 − P
E A′ ∩ E B′ ∩ E′ C
1 × P {EB }
1 − P
E A′
P
E′ B
P
E C′
P {EB }
1 − (1 − P {EA})(1 − P {EB })(1 − P {EC })
Similarly,
P {EC |E} =
P {EC }
1 − (1 − P {EA})(1 − P {EB })(1 − P {EC })
Using the notation above:
d(i) =
i
- 01 i(0.99)^20 −i
h(k) =
∑^20
i=k
d(i)
f (k) =
∑^20
i=k
i · P (i ICs fail to function properly | set fails to function properly ) =
∑^20
i=k
i ·
d(i)
h(k)
g(k) =
0 , if k = 1, ∑k− 1 i=1 i^ ·^ P^ (i^ ICs fail to function properly^ |^ set functions properly)
0 , if k = 1, ∑k− 1 i=1 i^ ·^
P ( set functions properly|i ICs fail to function properly )∗P (i ICs fail to function properly ) P ( set functions properly)
0 , if k = 1, ∑k− 1 i=1 i^ ·^
d(i) 1 −h(k) ,^ if^ k >^1
Let:
χ number of failed ICs in the product
χA number of failed ICs in set A
χB number of failed ICs in set B
χC number of failed ICs in set C
P (E) = 1 − (1 − h(1))(1 − h(m))(1 − h(2))
P (EB |E) =
P (E|EB ) · P (EB )
P (E)
h(m)
P (E)
E[χB |E] = E[χB |EB ] ∗ P (EB |E) + E[χB |EB ] ∗ (1 − P (EB |E)) =
f (m) · h(m)
P (E)
h(m)
P (E)
Similarly,
E[χA|E] =
f (1) · h(1)
P (E)
h(1)
P (E)
E[χC |E] =
f (2) · h(2)
P (E)
h(2)
P (E)
Finally,
E[χ|E] = E[χA|E] + E[χB |E] + E[χC |E] =
h(1)(f (1) − g(1)) + h(m)(f (m) − g(m)) + h(2)(f (2) − g(2))
P (E)
h(1)(f (1) − g(1)) + h(m)(f (m) − g(m)) + h(2)(f (2) − g(2))
1 − (1 − h(1))(1 − h(m))(1 − h(2))
