STAT500 HW#2_solutions
Solutions to Homework 2
1) (fifth)
a)
A={6}
n(A) / n(S) = 1/6
b)
B={2,4,6}
n(B) / n(S) = 3/6=1/2
c)
C={3,4,5,6}
n(C) / n(S) = 4/6=2/3
d)
D={4,6}
n(D) / (S)=2/6=1/3
(sixth)
a)
A={6}
n(A) / n(S) = 1/6
b)
B={1,3,5}
n(B) / n(S) = 3/6=1/2
c)
C={4,5,6}
n(C) / n(S) = 3/6=1/2
d)
D={4,6}
n(D) / (S)=2/6=1/3
2)
X∩Y
Mutually exclusive?
X∩Y =Ø?
P(X∩Y)
P(X)P(Y)
Independent?
P(X∩Y)=P(X)P(Y)?
A∩B
{6}
X
1/6
(1/6)(3/6)
X
A∩C
{6}
X
1/6
(1/6)(4/6)
X
A∩D
{6}
X
1/6
(1/6)(2/6)
X
B∩C
{4,6}
X
2/6=1/3
(3/6)(4/6)=1/3
O
B∩D
{4.6}
X
2/6=1/3
(3/6)(1/3)=1/6
X
C∩D
{4,6}
X
2/6=1/3
(4/6)(2/6)=2/9
X
Event B and event C are independent.
(OR, Event A, B, and C are not independent,
because P(A∩B∩C) = P({6})=1/
6
1/18 = P(A)*P(B)*P(C). )
≠
None of the pairs have empty intersections, therefore none of the events are mutually
exclusive.
3)
a)
P(A)=0.20+0.15+0.30=0.65,
P(B)=0.10+0.30=0.40.
b)
P(both A and B occur) = P(A∩B) =P({5})=0.30
c)
P(either A or B occurs) = P(A
U
B)=P({1,3,4,5})=0.20+0.15+0.10+0.30=0.75
d)
P(A∩B)=0.30 ≠ 0.26=0.65*0.40=P(A)P(B),
A and B are not independent.
e)
A∩B={5}≠Ø,
A and B are not mutually exclusive.
4)
a)
P (professional would accept promotion) = 0.7435.
P (professional rejects promotion) = 1

0.7435 = 0.2565.
b)
P (professional is part of twocareer marriage) = 0.46.
P (professional is part of onecareer marriage) = 0.37.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentSTAT500 HW#2_solutions
5)
First let's list the events and probabilities.
Let event A = pay in full any month (or first
month) and let B = pay in full the next month (or second month).
Then we have the
following probabilities:
This is the end of the preview.
Sign up
to
access the rest of the document.
 '08
 Chow
 Normal Distribution, π, BA, A∩C A∩D B∩C, professional rejects promotion, possible event outcomes

Click to edit the document details