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Home  >>  CBSE XII  >>  Math  >>  Probability
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An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $ \text {P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15}$. Evaluate the following probabilities \[ \text {(i) P(A fails | B has failed) (ii) P(A fails alone)} \]

$\begin{array}{1 1}(i) 0.5 (ii) 0.05 \\ (i) 0.05 (ii) 0.05 \\(i) 0.05 (ii) 0.5 \\(i) 0.5 (ii) 0.5 \end{array} $

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  • $P(E/F)=\large\;\frac{p(E\;\cap\;F)}{p(F)}$
Given P(A fails) = 0.2 P(B fails alone) = 0.15 P(A' $\cap$ B') = 0.15.
P (B fails alone) = 0.15 = P (B') - P (A' $\cap$ B') $\rightarrow$ 0.15 = p (B') - 0.15 $\rightarrow$ P (B') = 0.3.
(i) P (A fails | B has failed) $= \large\frac{P (A' \cap B')}{P(B')} = \large\frac{0;15}{0.30} $$= 0.5$
(ii) P (A fails alone) = P (A fails) - P (both A and B fail) = P (A') - P (A' $\cap$ \ B') = 0.2 - 0.15 = 0.05
answered Jun 22, 2013 by balaji.thirumalai
 

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