2010 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested Solutions

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

(i)
Required Probability
= \frac{3 \times 4 \times 3 \times 2 \times 1}{5!}

= \frac{3}{5}

(ii)
Required Probability
= \frac{3 \times 2 \times 1 \times 2 \times 1}{5!}

= \frac{1}{10}

(iii)
Case 1: The first digit is 3 or 5

\frac{2 \times 3 \times 2 \times 1 \times 2}{5!}

= \frac{1}{5}

Case 2: The first digit is 4

\frac{1 \times 3 \times 2 \times 1 \times 3}{5!}

= \frac{3}{20}

Required Probability = \frac{1}{5} + \frac{3}{20} = \frac{7}{20}

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