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 $= \frac{37}{300} \times \frac{36}{299} = 0.0148$