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

1. P(A fails| B has failed)
2. P(A fails alone)

Answer 1

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