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ST 351
Midterm 2 solutions – 70 points
Fall 2005
Form 2
Part I
:
1.
a
6.
d
9.
a
11.
c
(2.6 x 0.25)
13. a
15. b
16. a
2.
a
7.
d
10.
a
12.
c
14. d
(0.3 x 0.4 x 0.3)
3.
b
8.
d
(three ways one child could get the disease. Each has the same probability: (.2)(.8)(.8) =
4.
a
0.128. Multiply this probability by 3 to take into account the 3 different ways of one child
5.
b
getting the disease: 3(0.128) = 0.384)
Part II:
1. a.
No.
Multiple ways to show this – here’s one:
Let A = person snorkeled and let B = person is an adult.
P(A) = 33/94 = .3511.
P(B) = 61/94 = .6489.
P(A and B) = 25/94 = .26596.
P(A) x P(B) = (.3511)(.6489) = .2278 T .26596. Therefore, “activity at the beach” and “age” are
NOT independent.
b.
P(surf  youth) =
P(surf and youth)
7/94
=
P(youth)
33/94
=
.2121
2. First three do NOT hit the gold spot and the fourth does. Therefore, find P(1
st
no gold AND 2
nd
no gold
AND 3
rd
no gold AND 4
th
gold) = P(no gold) x P(no gold) x P(no gold) x P(gold) (since independent)
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 Spring '05
 KOLLATH

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