probability Q2 A study was carried out to investigate the causes of power failur
ID: 3181982 • Letter: P
Question
probability Q2
Explanation / Answer
From information given, P(power failure due to transformer damage)=0.10, P(power failure due to line problem)=0.75, and P(power failure due to transformer and line problem)=0.05
a. P(power failure due to transformer damage| line problem)=P(power failure due to line problem and transformer damage)/P(line problem)=0.05/0.75 [using rule for conditional probability, P(B|A)=P(A and B)/P(A)]
=0.0667 (ans)
b. P(line problem and transformer damage')=0.75*(1-0.10) [using general multiplication rule, P(A and B)=P(A)*P(B), P(B')=1-P(B)]
=0.675 (ans)
c. P(transformer damage or line problem)=P(power failure due to transformer damage)+P(power failure due to line problem)=0.10+0.75 [using general addition rule, P(A or B)=P(A)+P(B)]
=0.85 (ans)
Declare two events independent, if P(power failure due to transfomer damage|line damage)=P(power failure due to transformer problem)
Here, P(power failure due to transformer damage| line problem)=0.0667 and P(power failure due to transformer damage)=0.10
Because, 0.6667=/=0.10, the events are not independent.
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