Chapter 4:
Probability
1
Chapter 4
Probability
Ans Not Included is 4.30
4.1
Enumeration of the six parts:
D
1
, D
2
, D
3
, A
4
, A
5
, A
6
D = Defective part
A = Acceptable part
Sample Space:
D
1
D
2
,
D
2
D
3
,
D
3
A
5
D
1
D
3
,
D
2
A
4
,
D
3
A
6
D
1
A
4
,
D
2
A
5
,
A
4
A
5
D
1
A
5
,
D
2
A
6
,
A
4
A
6
D
1
A
6
,
D
3
A
4
,
A
5
A
6
There are 15 members of the sample space
The probability of selecting exactly one defect out of two is:
9/15 =
.60
4.2
X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9} and Z = {1, 2, 3, 4, 7,}
a)
X
⊥
Z =
{1, 2, 3, 4, 5, 7, 8, 9}
b)
X
_
Y =
{7, 9}
c)
X
_
Z =
{1, 3, 7}
d)
X
⊥
Y
⊥
Z =
{1, 2, 3, 4, 5, 7, 8, 9}
e)
X
_
Y
_
Z =
{7}
f)
(X
⊥
Y)
_
Z = {1, 2, 3, 4, 5, 7, 8, 9}
_
{1, 2, 3, 4, 7} =
{1, 2, 3, 4, 7}
g)
(Y
_
Z)
⊥
(X
_
Y) = {2, 4, 7}
⊥
{7, 9} =
{2, 4, 7, 9}
h)
X or Y = X
⊥
Y =
{1, 2, 3, 4, 5, 7, 8, 9}
i)
Y and Z = Y
_
Z =
{2, 4, 7}
4.3
If A = {2, 6, 12, 24} and the population is the positive even numbers through 30,
A’ =
{4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}
4.4
6(4)(3)(3) =
216
4.5
Enumeration of the six parts:
D
1
, D
2
, A
1
, A
2
, A
3
, A
4
D = Defective part
A = Acceptable part
Sample Space:
D
1
D
2
A
1
,
D
1
D
2
A
2
,
D
1
D
2
A
3
,
D
1
D
2
A
4
,
D
1
A
1
A
2
,
D
1
A
1
A
3
,
D
1
A
1
A
4
,
D
1
A
2
A
3
,
D
1
A
2
A
4
,
D
1
A
3
A
4
,
D
2
A
1
A
2
,
D
2
A
1
A
3
,
D
2
A
1
A
4
,
D
2
A
2
A
3
,
D
2
A
2
A
4
,
D
2
A
3
A
4
,
A
1
A
2
A
3
,
A
1
A
2
A
4
,
A
1
A
3
A
4
,
A
2
A
3
A
4
Combinations are used to counting the sample space because sampling is done without replacement.
6
C
3
=
!
3
!
3
!
6
= 20
Probability that one of three is defective is:
12/20 = 3/5
.60
There are 20 members of the sample space and 12 of them have 1 defective part.
4.6
10
7
= 10,000,000 different numbers
4.7
20
C
6
=
!
14
!
6
!
20
=
38,760
It is assumed here that 6 different (without replacement) employees are to be selected.
4.8
P(A) = .10, P(B) = .12, P(C) = .21
P(A
_
C) = .05
P(B
⊥
C) = .03
a) P(A
⊥
C) = P(A) + P(C) - P(A
_
C) = .10 + .21 - .05 =
.26
b) P(B
⊥
C) = P(B) + P(C) - P(B
_
C) = .12 + .21 - .03 =
.30
c) If A, B mutually exclusive,
P(A
⊥
B) = P(A) + P(B) = .10 + .12 =
.22
4.9
D
E
F
A
5
8
12
25
B
10
6
4
20
C
8
2
5
15
23
16
21
60
a) P(A
⊥
D) = P(A) + P(D) - P(A
_
D) =
25/60 + 23/60 - 5/60 = 43/60 =
.7167
b) P(E
⊥
B) = P(E) + P(B) - P(E
_
B) =
16/60 + 20/60 - 6/60 = 30/60 =
.5000
c) P(D
⊥
E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 =
.6500
d) P(C
⊥
F) = P(C) + P(F) - P(C
_
F) =
15/60 + 21/60 - 5/60 = 31/60 =
.5167