31
Chapter 3: Discrete Random Variables and Their Probability Distributions
3.1
P
(
Y
= 0) =
P
(no impurities) = .2,
P
(
Y
= 1) =
P
(exactly one impurity) = .7,
P
(
Y
= 2) = .1.
3.2
We know that
P
(
HH
) =
P
(
TT
) =
P
(
HT
) =
P
(
TH
) = 0.25.
So,
P
(
Y
= 1) = .5,
P
(
Y
= 1) =
.25 =
P
(
Y
= 2).
3.3
p
(2) =
P
(
DD
) = 1/6,
p
(3) =
P
(
DGD
) +
P
(
GDD
) = 2(2/4)(2/3)(1/2) = 2/6,
p
(4) =
P
(
GGDD
) +
P
(
DGGD
) +
P
(
GDGD
) = 3(2/4)(1/3)(2/2) = 1/2.
3.4
Define the events:
A
: value 1 fails
B
: valve 2 fails
C
: valve 3 fails
)
(
)
2
(
C
B
A
P
Y
P
∩
∩
=
=
= .8
3
= 0.512
)
(
)
(
))
(
(
)
0
(
C
B
P
A
P
C
B
A
P
Y
P
∪
=
∪
∩
=
=
= .2(.2 + .2  .2
2
) = 0.072.
Thus,
P
(
Y
= 1) = 1  .512  .072 = 0.416.
3.5
There are 3! = 6 possible ways to assign the words to the pictures.
Of these, one is a
perfect match, three have one match, and two have zero matches.
Thus,
p
(0) = 2/6,
p
(1) = 3/6,
p
(3) = 1/6.
3.6
There are
2
5
= 10 sample points, and all are equally likely: (1,2), (1,3), (1,4), (1,5),
(2,3), (2,4), (2,5), (3,4), (3,5), (4,5).
a.
p
(2) = .1,
p
(3) = .2,
p
(4) = .3,
p
(5) = .4.
b.
p
(3) = .1,
p
(4) = .1,
p
(5) = .2,
p
(6) = .2,
p
(7) = .2,
p
(8) = .1,
p
(9) = .1.
3.7
There are 3
3
= 27 ways to place the three balls into the three bowls.
Let Y = # of empty
bowls.
Then:
p
(0) =
P
(no bowls are empty) =
27
6
27
!
3
=
p
(2) = P(2 bowls are empty) =
27
3
p
(1) = P(1 bowl is empty) = 1
27
18
27
3
27
6
=
−
−
.
3.8
Note that the number of cells cannot be odd.
p
(0) =
P
(no cells in the next generation) =
P
(the first cell dies or the first cell
splits and both die) = .1 + .9(.1)(.1) = 0.109
p
(4) =
P
(four cells in the next generation) = P(the first cell splits and both created
cells split) = .9(.9)(.9) = 0.729.
p
(2) = 1 – .109 – .729 = 0.162.
3.9
The random variable
Y
takes on vales 0, 1, 2, and 3.
a.
Let
E
denote an error on a single entry and let
N
denote no error.
There are 8 sample
points:
EEE
,
EEN
,
ENE
,
NEE
,
ENN
,
NEN
,
NNE
,
NNN
.
With
P
(
E
) = .05 and
P
(
N
) = .95
and assuming independence:
P
(
Y
= 3) = (.05)
3
= 0.000125
P
(
Y
= 2) = 3(.05)2(.95) = 0.007125
P
(
Y
= 1) = 3(.05)
2
(.95) = 0.135375
P
(
Y
= 0) = (.95)
3
= 0.857375.
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Chapter 3: Discrete Random Variables and Their Probability Distributions
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