All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{P}(B|A) = \mathrm{P}(B)$ since A and B are independent.

$\therefore, \mathrm{P}(B|A) = \mathrm{P}(B) = 0.4$

(ii)
$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(C \cup B \cup A)$

$= 1 - (0.45 + 0.02 + 0.2 + 0.145)$

$= 0.185$

(iii)

If $\mathrm{P}(A' \cap B' \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$=1 - [0.935 - (0.3 - 0.035 )]$

$= 0.33$ (maximum)

If $\mathrm{P}(A' \cap B \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(A \cup B \cup C) - 0.22$

$= 0.165$ (minimum)

$0.165 \le \mathrm{P}(A' \cap B' \cap C') \le 0.33$