2002 A-level H2 Mathematics Paper 2 Question 30 Suggested Solutions

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

(a)
(i)
$\frac{7 \times 6 \times 5}{7 \times 7 \times 7} = \frac{30}{49}$
Alternatively, $\frac{^7 P_3}{7^3}$

(ii)
When $n=4$ , required probability $= \frac{^7 P_4}{7^3} = \frac{7 \times 6 \times 5 \times 4}{7 \times 7 \times 7 \times 7} = \frac{120}{343}$

(b)
Key GC with either sequence or use table functions.
We want to solve $\frac{^12 C_n \times n!}{12^n} \textless \frac{1}{2}$
Plot $y_1 = \frac{^12 C_x \times x!}{12^x}$ and check for the x that gives a corresponding $y_1$ value that is less than $\frac{1}{2}$

Probability that all 21 people have different birthdays $= \frac{^365 P_23}{365^23} = 0.4927$

Thus, when $n =23$ , probability that at least two of the people are the same birthday $= 1 - 0.4927 = 0.5073$ (more than $\frac{1}{2}$ )

KS Comments:

A very interesting probability question here. This show show counter intuitive probability is! The results shows that the we can expect to find someone with the same birthday in a room of 23, more than half the time.

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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.

2002 A-level H2 Mathematics Paper 2 Question 30 Suggested Solutions

KS Comments:

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