Chapter 5
2.
a.
y
p(x,y)
0
1
2
3
4
0
.30
.05
.025
.025
.10
.5
x
1
.18
.03
.015
.015
.06
.3
2
.12
.02
.01
.01
.04
.2
.6
.1
.05
.05
.2
b.
P(X
1 and Y
1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .56
= (.8)(.7) = P(X
1)
P(Y
1)
c.
P( X + Y = 0) = P(X = 0 and Y = 0) = p(0,0) = .30
d.
P(X + Y
1) = p(0,0) + p(0,1) + p(1,0) = .53
7.
e.
p(1,1) = .030
f.
P(X
1 and Y
1 = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .120
g.
P(X = 1) = p(1,0) + p(1,1) + p(1,2) = .100; P(Y = 1) = p(0,1) + … + p(5,1) = .300
h.
P(overflow) = P(X + 3Y > 5) = 1
–
P(X + 3Y
5) = 1
–
P*(X,Y)=(0,0) or …or (5,0)
or (0,1)
or (1,1) or (2,1)] = 1 - .620 = .380
i.
The marginal probabilities for X (row sums from the joint probability table) are p
x
(0)
= .05, p
x
(1) = .10 , p
x
(2) = .25,
p
x
(3) = .30, p
x
(4) = .20, p
x
(5) = .10; those for Y (column
sums) are p
y
(0) = .5, p
y
(1) = .3, p
y
(2) = .2.
It is now easily verified that for every (x,y),
p(x,y) = p
x
(x)
p
y
(y), so X and Y are independent.