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}

(c)
When n =21, probability that all 21 people have different birthdays = \frac{^365 P_21}{365^21} = 0.55631

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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