Question:
An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15 Evaluate the following probabilities (i) P(A fails|B has failed) (ii) P(A fails alone)
Answer:
Let EA denotes the events in which A fails.
and Let EB denotes the events in which B fails.
Given P(EA ) = 0.2
and P(EA ∩ EB ) = 0.15
P(B fails alone) = P(EB ) - P(EA ∩ EB )
=> 0.15 = P(EB ) - 0.15
=> P(EB ) = 0.15 + 0.15
=> P(EB ) = 0.30
1. P(EA /EB ) = P(EA ∩ EB )/P(EB ) = 0.15/0.3 = 0.5
2. P(A fails alone) = P(EA ) - P(EA ∩ EB )
= 0.2 - 0.15
= 0.05
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