Solutions to Homework #5
(W4150 ‘Intro to Probability and Statistics’, S04)
Sec. 5.3. (4)
(a) P(X=2) = C
5
2
(3/4)
2
(1/4)
3
= .0879 ; or you can use tables/Excel to find the same value by
taking difference P(X=2) = P(X<
2) – P(X<
1) =
2
0
b(x ;5,.75) 
1
0
b(x ;5,.75)
(b) P(X<
3) =
3
0
b(x ;5,.75) = .3672
Sec.5.3. (7)
(a) p = .7; n = 10
±
P(X<n/2) = P(X<5) = P(X<
4) =
4
0
b(x ;10,.7) = .0473
(b) p = .7; n = 20
±
P(X<n/2) = P(X<10) = P(X<
9) =
9
0
b(x ;20,.7) = .0171
Sec. 5.3. (15)
P(inoculated mouse is protected) = p = .6; n = 5. Let X = {# of mice that didn’t get ill}
(a) P(none of the mice is infected} = P(X = n = 5) =
5
0
b(x ;5,.6) 
4
0
b(x ;5,.6) = .0778
(b) P(fewer than 2 contract the disease) = P(X>
4) = 1 – P(X<4) = 1 
3
0
b(x ;5,.6) = .337
(c) P(more than 3 contract the disease) = P(X<
1) =
1
0
b(x ;5,.6) = .087
Comment
: if you choose X to be number of ill mices than you need to preserve «connection»
between the notion of ‘success’ and it’s probability. Namely, if you adopt the above
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 Spring '04
 GALLEGO
 Poisson Distribution, Probability, Binomial distribution, Sec., inoculated mouse

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