2013 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

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

(i)

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

(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.

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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2013 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

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