Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 20
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20Answers
13. Let D denote the event that the chosen machine is defective and ¯ D denote the event
“not D”.
Let Y be the number of imperfect articles in the sample of five.
Then
P(D)
P(Y = 2 D)
P(D Y = 2) =
P(Y = 2 D) + P( ¯ D)
¯ D)
P(D)
P(Y = 2
5
2
2
3
0.2
0.8
5
2
=
5
5
2
3
2
3
2
3
0.2
0.8
+
0.1
0.9
5
5
2
2
2
3
2
0.2
0.8
=
2
3
2
3
2
0.2
0.8
+ 3
0.1
0.9
0.04096
=
= 0.6519.
0.04096 + 0.02187
14.
20
19
18
20
3
17
3
7
(a) (i)
p
=
0.1
0.9
=
0.1
0.9
= 0.190.
3
3
1
2
3
(ii)
3
20
2
18
p
=
0.1
0.9
=
9
p
= 0.28518
2
3
2
18
2
20
19
p
=
0.1
0.9
=
9
p
= 0.27017
1
2
1
19
20
20
p
=
0.9
= 0.12158.
0
0
The total probability is 0.867.
(iii) The required probability is the probability of at most 2 out of 16.
16
p
= P(0 out of 16) = 0.9
= 0.185302
0
16
p
= P(1 out of 16) =
p
= 0.3294258
1
0
9
15
1
p
= P(2 out of 16) =
p
= 0.2745215
2
1
2
9
(b)
4
1
3
0.2
0.3
0.7
1
0.02058
=
= 0.2608.
0.02058 + 0.05832
4
4
1
3
1
3
0.2
0.3
0.7
+ 0.9
0.1
0.9
1
1
36
HELM (2008):
Workbook 37: Discrete Probability Distributions
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