2014 A-level H1 Mathematics (8864) Question 11 Suggested Solutions

By KS Teng

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{100}{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.

Showing 2 comments
  • SomeGuy
    Reply

    Typo in question 1, as your equation writes 10/(290+x) instead of 100/(290+x)

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