Probability Assignment 4: Decision Making & Bayesian Analysis

This document presents the solution to Probability Assignment 4, focusing on decision-making under uncertainty and Bayesian analysis. The first question explores optimal actions based on different probability scenarios, calculating posterior expected losses to determine the best course of action. The second question applies probability to a real-world problem of weld inspection, evaluating the cost-effectiveness of a new sonic device. The solution calculates the probability of finding a bad weld indication and compares the total cost incurred by accepting the new method versus the cost of defective welds leaving the shop, ultimately determining whether the new device should be accepted. Desklib offers a variety of solved assignments and past papers for students.

PROBABILITY
ASSIGNMENT 4
Student Name
[Pick the date]

Question 1
Charging 20 per hour or 25 per hour
Case 1:
Fee level a1= 25 per hour
Business response s1 =Adequate
Gain = 100
Business response s2= Satisfactory
Loss = 50
Case 2:
Fee level a2= 20 per hour
Business response s1 =Adequate
Gain = 50
Business response s2= Unsatisfactory
Loss = 100
(a) Optimum action if P ( s1|a1 ¿=0.5∧P ( s2| a2 ¿=0.5
The optimum action can be determined with the help of Bayer’s action that minimizes the
posterior expected loss.
It can be seen that loss is not a negative value and therefore,
a1 a2
s1 −100 −50
s2 50 100
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Now, posterior probabilities can be determined as shown below:
P ( s1|a1 ¿=0.5
P ( s2|a1 ¿=1−0.5=0.5
P ( s2|a2 ¿=0.5
P ( s1|a2 ¿=0.5
The posterior expected losses for a1is calculated below:
E ( L ( s2 a )|a1 ¿=L ( s1 a1 ) . P ( s1| a1 ¿+ L ( s2 a1 ) . P ( s2| a1 ¿
E ( L ( s2 a )|a1 ¿= { (−100 )∗( 0.5 ) }+{ (50 )∗( 0.5 ) }
E ( L ( s2 a )|a1 ¿=−25
The posterior expected losses for a2is calculated below:
E ( L ( s1 a )|a2 ¿=L ( s1 a2 ) . P (s2|a2 ¿+ L ( s2 a2 ) . P ( s2|a2 ¿
E ( L ( s1 a )|a2 ¿= { (−50 )∗( 0.5 ) }+ {( 100 )∗( 0.5 ) }
E ( L ( s1 a )|a2 ¿=25
Therefore, the optimum action is a1 because E ( L ( s2 a )|a1 ¿< E ( L ( s1 a )|a2 ¿
(b) Optimum action if P ( s1|a1 ¿=0.3∧P ( s2| a2 ¿=0.4
P ( s1|a1 ¿=0.3
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P ( s2|a1 ¿=1−0.3=0.7
P ( s2|a2 ¿=0.4
P ( s1|a2 ¿=1−0.4=0.6
The posterior expected losses for a1is calculated below:
E ( L ( s1 a )|a1 ¿= { ( −100 ) ∗( 0.3 ) } +{ ( 50 )∗( 0.7 ) }
E ( L ( s1 a )|a1 ¿=5
The posterior expected losses for a2is calculated below:
E ( L ( s1 a )|a2 ¿= { (−50 )∗( 0.6 ) }+ { (100 )∗( 0.4 ) }
E ( L ( s1 a )|a2 ¿=10
Therefore, the optimum action is a1 because E ( L ( s1 a )|a1 ¿< E ( L ( s1 a )| a2 ¿
Question 2
Probability that in new device a flaw is present = 0.80
Probability that new device erroneously pass the weld = 0.20
Probability that new device will pass specimen and no flaw is present =0.90
Probability that new device indirectly indicates critical flaw = 0.10
Percentage of welds are rejected and rewelded = 15%
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(a) Probability to find an indication of bad weld with the help of sonic device.
P ( Indication of finding bad weld )=P ( Rejected weld )∗P ( Indicates that flaw is pressent )+ P ( Accepted weld )∗P( In
P ( Indication of finding bad weld ) = ( 0.15∗0.80 ) +(0.85∗0.10)
P ( Indication of finding bad weld ) =0.205
(b) The aim is to determine whether or not to accept the new device.
Inspection cost = $1.00 per weld
New method costs = $0.50 per weld
Cost to redo the weld = 5
$30 to have a defective weld leaves the shop.
Now,
Total cost incurred when one can accept the new method = P (New method) + {P (Redo a correct
weld) * $5} + { P (a defective weld will leave the shop)* $30}
Here,
P (Redo a correct weld) = P (False positive) = 0.10 *0.85 = 0.085
Similarly,
P (a defective weld will leave the shop) = 0.15 * 0.20 = 0.03
Hence,
Total cost incurred when one can accept the new method = P (New method) + {P (Redo a correct
weld) * $5} + {P (a defective weld will leave the shop)* $30}
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Total cost incurred when one can accept the new method = 0.50 + (0.085 * $5) + (0.03* $30) =
$1.825
Based on the above, it can be said that new device does not to be accepted.
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