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

(i)
(a) Required Probability = (0.1)(0.2)(0.1) = 0.002

(b) Required Probability
= 1 - \mathrm{P(no~stars)}
= 1 - (0.9)(0.8)(0.9) = 0.352

(c)
\mathrm{P}(\times \times +) + \mathrm{P}(\times + \times) + \mathrm{P}(+ \times \times )
= (0.3)(0.1)(0.2) + (0.3)(0.3)(0.4) + (0.4)(0.1)(0.4)
= 0.058

(ii)
\mathrm{P}(exactly~ 1 ~star)
= (0.1)(0.8)(0.9) + (0.2)(0.9)(0.9) + (0.1)(0.9)(0.8)
=0.306

\mathrm{P}(+,\circ~|~ exactly ~1 ~star)

=\frac{\mathrm{P}(+,\circ, \star)}{\mathrm{P}(exactly 1 \star)}

=\frac{(0.4)(0.4)(0.1)+(0.4)(0.2)(0.3)+(0.2)(0.3)(0.1)+(0.2)(0.2)(0.2)+(0.1)(0.3)(0.3)+(0.1)(0.4)(0.2)}{0.306}

= \frac{71}{306}

Personal Comments:
This question was a question that a few students of mine was jokingly telling me they see stars here. To be honest, this is not super hard. Just extremely tedious.

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