$\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} $

- $P(E/F)=\large\;\frac{p(E\;\cap\;F)}{p(F)}$

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

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