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I am really stuck at some probability exercise. Given that the system fails, what is the probability that the component B1 fails? The circuit is as follows:

Circuit

The probability that the system fails is: P=A'*B' = 0.01*0.02 = 0.0297. The probability that A fails is P(A'|System failure) = 0.01/0.0297 = 0,332

I can't figure out how to get the probability that B1 fails given that the system fails. Thanks in advance.

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    Something's wrong. In order for Sys B to fail, _both_ B1 and B2 (in parallel) have to fail. – 2017-02-13
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    I know that the probability that B works is 0.98 so that B fails must be 1 - P(B works) that gives us: 1 - 0.98 = 0.02.? – 2017-02-13
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    You're in a tangle. New approach. Make a Venn Diagram with A, B1, B2. Shade in _exactly_ the areas that lead to system failure. For exmp, Syst will fail of all 3 components fail. What else? Maybe you're not thinking correctly about the difference btw parallel and series connections. [Erasing one of my earlier comments, so we won't get busted for 'chatting' in Comments]. – 2017-02-13
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    I really don't get it. Do I have to take away something in my equation? – 2017-02-13

1 Answers 1

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The system fails if either subsystem $A$ fails or subsystem $B$ fails,
i.e. $P(\text{system fails})=A' + B1'B2' - A'B1'B2'$

Since each component failure is independent, P($B' \cap \text{system fails} ) = B1'A' + B1'B2'A$

$P(B1' |\text{system fails}) = \dfrac{P(B1'\cap \text{system fails})}{P(\text{system fails})} = \dfrac{0.1\times0.01 + 0.1\times0.2\times0.99}{0.01 + 0.1\times0.2- 0.01\times0.1\times0.2} =\dfrac{2.08}{2.98}$

Added list of cases where system fails (lower case represents failure of a component)

a b1 B2
a b1b2
a B1 B2
a B1 b2
A b1b2

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    Quibble: $P(Sys fail) = P(A') + P(B_1' \cap B_2') - P(A'\cap B_1'\cap B_2').$ Anyhow, right track (+1) – 2017-02-13
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    I'll add a list (there are only $5$ cases ! It'll be clear then. – 2017-02-13
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    In case anyone cares: my simulation with $10^8$ iterations gives 0.698, which agrees with my (pre-morning-coffee) computation. – 2017-02-13
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    There are $2^3 = 8$ cases with probabilities that add to 1. Look at Venn Diagram. – 2017-02-13
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    @BruceET: You are right, but for the question I have considered the restricted sample space where system failure occurs. – 2017-02-13
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    @PureRushh: If we take all $8$ cases, we'll get a probability of 100%, but I have taken only the $5$ cases that lead to system failure. – 2017-02-13
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    @trueblueanil Ahh so we are missing A B1 b2, A b1 B2, A B1 B2? – 2017-02-13
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    Yes, but I have spotted an obvious error, thanks to @BruceET (it's 1:30 am here) so hopefully I'll get it right now, and go to sleep ! – 2017-02-13