2013 A-level H2 Mathematics (9740) Paper 2 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)

Required Probability = \frac{26 \times 25 \times 24 \times 9 \times 8}{26 \times 26 \times 26 \times 9 \times 9} \approx 0.789

(ii)

Required Probability = \frac{1-\mathrm{P}(\mathrm{same digits})}{2}
= \frac{1-\frac{9}{9^{2}}}{2}
= \frac{4}{9}

(iii)
Case 1: 2 same letter and 2 different digits

26 \times 25 \times \frac{3!}{2!} \times {^9\!P_2}=140400

Case 2: different letter and 2 same digits

9 \times {^{26}\!P_3}=140400

Case 3:3 same letter and 2 same digits

9 \times 26 =234

Required Probability = \frac{140400+140400+234}{26 \times 26 \times 26 \times 9 \times 9} \approx 0.197

(iv)
Case 1: 2 different consonants, 1 vowel & 1 even digit

{^{21}\!C_2} \times {^5\!C_1} \times 3! \times {^4\!C_1} \times {^5\!C_1} \times 2! = 252000

Case 2: 2 same consonant, 1 vowel & 1 even digit

{^{21}\!C_1} \times {^5\!C_1} \times \frac{3!}{2!} \times {^4\!C_1} \times {^5\!C_1} \times 2! = 12600

Required Probability = \frac{252000+12600}{26 \times 26 \times 26 \times 9 \times 9} \approx 0.186

KS Comments:

This question is not that easy. A lot of thinking is required; but that is the case with all permutation questions. I strongly suggest students to not spend too much time on permutation questions and come back to solve it later.

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