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

(i)

\mathrm{P} (L) = \frac{90}{290+x}

\mathrm{P} (G) = \frac{100}{290+x}

Since L and G are independent events,

\mathrm{P} (L \cap G) = \mathrm{L} \times \mathrm{G}

\frac{30}{290+x} = (\frac{90}{290+x})(\frac{10}{290+x})

x = 10

(ii)
\mathrm{P} (L \cup T) = \frac{235}{300} = \frac{47}{60}

(iii)

\mathrm{P} (T \cap G) = \frac{142}{300} = \frac{71}{150}

(iv)

\mathrm{P} (L | G) = \mathrm{P}(L) since L and G are independent events.

\therefore, \mathrm{P} (L | G) =  \frac{90}{300} = \frac{3}{10}

(v)

Required Probability = \frac{37}{300} \times \frac{36}{299} = 0.0148

KS Comments

Some students commented that (v) should be done via binomial distribution, but this is a misguided approach as they should realise that the probability is not constant. Thus, a binomial model will not be appropriate.

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