1. Given
Information
-
They are independent to each other
-
0.05% defect rate at receiving
inspection
-
0.01% defect rate at other inspection
points
-
We take 50 samples at receiving
inspection and accept/reject rate is 0/1
-
We take 10 samples at any other
inspection points and accept/reject rate is 0/1
-
Find probability that purchased material
will make it through process with no reject
-
Using Binomdist function from Excel, ex.
=BINOMDIST(D3,B3,C3,TRUE), we get
|
Inspection Points
|
Sample Size
|
Defect Rate
|
Allowed Failure
|
Probability of passing
|
|
Receiving Inspection
|
50
|
0.0005
|
0
|
0.97530
|
|
Inspection 1
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 2
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 3
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 4
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 5
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 6
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 7
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 8
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 9
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 10
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 11
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 12
|
10
|
0.001
|
0
|
0.99004
|
|
Inspection 13
|
10
|
0.001
|
0
|
0.99004
|
|
0.85635
|
This table implies that even though the reliability
at individual inspection point is relatively high, when we combine all of the
reliability of the system, it reduces our system reliability significantly. For
our numbers, we have over 99% probability of passing inspection at
13 inspection points and 97.5% at the receiving inspection point, but
when we cumulative its probability by multiplying it, it declines significantly.
1. Chapter
2 problem 1
-
In the test firing of a missile, there
are some events that are known to cause the missile to fail to reach its
target. These events are listed below; together with their approximate
probabilities of occurrence during a flight:
Cloud
Reflection=0.0001
Precipitation=0.005
Target
Evasion=0.002
Electronic
countermeasures=0.04
The
probabilities of failure if these events occur are:
P(F/Cloud
Refelction)=0.3;P(F/Precipitation)=0.01,P(F/Target
Evasion)=0.005,P(F/Electronic Countermeasures)=0.0002.
Use
Bayes’ theorem to calculate the probability of each of these events being the
cause in the event of a missile failing to reach its target.
Using Bayes’ theorem, P(A|X) =
[P(A)P(X|A)]/[P(A)P(X|A) + P(B)P(X|B) + P(C)P(X|C)],
We get this result
|
Probability of Failing
|
Probability of Happening
|
Probability of it causing
failure
|
|
|
Cloud Reflection
|
0.3
|
0.0001
|
0.047021944
|
|
Precipitation
|
0.1
|
0.005
|
0.78369906
|
|
Target Evasion
|
0.05
|
0.002
|
0.156739812
|
|
Electronic
Countermeasures
|
0.0002
|
0.04
|
0.012539185
|
|
1
|
2.
Chapter 2 Problem 2
For a device with a failure probability of
0.02 when subjected to a specific test environment, use the binomial
distribution to calculate the probabilities that a test sample of 25 devices
will contain (a)no failure; (b)one failure; (c)more than one failure. Compute
the same problem with probability of 0.2 instead.
-
Using Excel to compute the calculation.
|
Sample Size
|
Failure Rate
|
Allowed Failure
|
Probability of Passing
|
|
25
|
0.02
|
0
|
0.60346473
|
|
25
|
0.02
|
1
|
0.911354898
|
|
25
|
0.02
|
2
|
0.986756572
|
|
25
|
0.02
|
3
|
0.998554113
|
|
25
|
0.02
|
4
|
0.999878326
|
|
25
|
0.02
|
5
|
0.999991831
|
|
Sample Size
|
Failure Rate
|
Allowed Failure
|
Probability of Passing
|
|
25
|
0.2
|
0
|
0.003777893
|
|
25
|
0.2
|
1
|
0.027389726
|
|
25
|
0.2
|
2
|
0.098225223
|
|
25
|
0.2
|
3
|
0.233993259
|
|
25
|
0.2
|
4
|
0.420674309
|
|
25
|
0.2
|
5
|
0.616689412
|
As we increase our failure rate to 20% from 2%, the
probability of passing decreases dramatically.
Probability of getting 1 defect is just probability of 1
defect – probability of 0 defect or we could just simply have false statement
on the excel command which gives us a mass function of the distribution. We get
the probability of 30.79% for getting 1 defect with 2% defect rate. We get2.36%
for getting 1 defect with 20% defect rate.
For defects more than 1, we simply subtract probability of
getting 1 defect from 1. We get 8.86%
for 2% defect rate and 97% for 20% defect rate.
3.
Chapter 2 Question 4
Repeat the question 2,3 with Poission
Distribution.
-
In order to do the computation with Poission
distribution, we need the mean. We simply calculate the mean by multiplying its
defect rate with the sample number.
Doing
the computation with excel, we get these results.
|
Sample Size
|
Failure Rate
|
Mean
|
Probability of Passing
|
|
25
|
0.02
|
0.5
|
0.60653066
|
|
25
|
0.02
|
0.5
|
0.90979599
|
|
25
|
0.02
|
0.5
|
0.985612322
|
|
25
|
0.02
|
0.5
|
0.998248377
|
|
25
|
0.02
|
0.5
|
0.999827884
|
|
25
|
0.02
|
0.5
|
0.999985835
|
|
Sample Size
|
Failure Rate
|
Allowed Failure
|
Probability of Passing
|
|
25
|
0.2
|
5
|
0.006737947
|
|
25
|
0.2
|
5
|
0.040427682
|
|
25
|
0.2
|
5
|
0.124652019
|
|
25
|
0.2
|
5
|
0.265025915
|
|
25
|
0.2
|
5
|
0.440493285
|
|
25
|
0.2
|
5
|
0.615960655
|
In comparison with results from binomial distribution
function with the poisson distribution, we see that the result probability is
relatively the same that we can use either of method to compute the interested
probability depends on the situation.
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