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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
- P(A fails| B has failed)
- P(A fails alone)
Solution
The events of failure A and B are `barA` and `barB` respectively.
∴ `P(barA)` = 0.2
and P(fail A and B) = 0.15
⇒ `P(barA ∩ barB)` = 0.15
`P("alone" barB) = P(barB) - P(barA ∩ barB) = 0.15`
Now, 0.15 = `P(barB) - 0.15`
∴ `P(barB) = 0.30`
i. `P(barA/barB) = (P(barA ∩ barB))/(P(barB))`
= `0.15/0.30`
= `1/2`
= 0.5
ii. P(A alone fails) = `P("alone" barA)`
= `P(barA) - P(barA ∩ barB)`
= 0.20 − 0.15
= 0.05
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