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Type I Problems
1.
#21, page 349 of the text, 9
th
edition.
E[Time] = E[Time waiting at 1] + E[Service at 1] + E[Time waiting at 2] + E[Service at 2]
E[Service at 1] = 1/μ
1
E[Service at 2] = 1/μ
2
E[Time waiting at 1] = 1/μ
1
since we just have to wait for the current customer at 1 to finish
E[Time waiting at 2] = 0*P(2 free when finish 1) + E[2 to finish]* P(2 not free when finish 1)
= 1/μ
2
(μ
1
/(μ
1
+μ
2
)) by the alarm clock lemma
So E[Time] = 2/μ
1
+ 1/μ
2
(1 + μ
1
/(μ
1
+μ
2
))
2.
#34, page 351 of the text, 9
th
edition.
Let A = time until A dies, B = time until B dies, T
i
= kidney waiting time (for i = kidney 1, 2)
a. P(A receives a kidney) = P(T < A)
= λ/( λ + μ
A
) by the alarm clock lemma
b. P(B receives a kidney) = P(B receives the first kidney) + P(B receives the second kidney)
= P(A < T
1
< B) +
P(T
1
< A and T
1
+ T
2
< B)
= P(A = min{T
1
, A, B})*P(T
1
= min{T
1
, B}) + P(T1 = min{T
1
, A, B})*P(B>T1+T2B>T1)
= (μ
A
/ (μ
A
+ μ
B
+
λ))*( λ / (μ
B
+
λ)) + (λ / (μ
A
+ μ
B
+
λ))* ( λ / (μ
B
+
λ))
using the alarm clock lemma and the memoryless property
= ( λ / (μ
B
+
λ))* ((μ
A
+ λ)
/ (μ
A
+ μ
B
+
λ))
3.
#59, page 356 of the text, 9
th
edition.
Let 4000 be the event of a $4000 claim, 1 be the event it was type 1, and 2 be the event it was
type 2. We want P(14000).
Using Bayes’ rule, P(14000) = P(40001)P(1) / [P(40001)P(1) + P(40002)P(2) ]
Since the Poisson processes for type 1 and 2 claims are independent, the process of all claims is
also Poisson but with rate 10 + 1 = 11. Thus P(1) = 10/11 and P(2) = 1/11.
The probability a claim is $4000 would theoretically be 0 since we have a continuous
distribution of claim size. However, we can think of a $4000 claim being a claim that is between
$4000.00 and $4000.01 (or any other similarly small interval.)
Thus P(40001) ≈ f
1
(4000)(0.01) = (1/1000)e
– 4000/1000
0.01 = (0.01/1000)e
– 4
And P(40002) ≈ f
2
(4000)(0.01) = (1/5000)e
– 4000/5000
0.01 = (0.01/5000)e
– 0.8
Plugging everything in, the interval length (0.01) cancels out. You could basically ignore the fact
that this was a continuous distribution and just use the pdf like a pmf… but I like to be precise.
We get P(14000) = e
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 Spring '08
 Chisholm

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