Discussion
In a year of 366 days, what is the chance that three persons have different birthdays?
Answer: C
The probability is \(\frac{365 \times 364}{366 \times 366}\).
Take the three people one at a time. The first can have any of the 366 days. The second must avoid that day, and the third must avoid both earlier days:
\[ \begin{aligned} P &= \frac{366}{366} \times \frac{365}{366} \times \frac{364}{366} \\ &= \frac{365 \times 364}{366 \times 366} \end{aligned} \]
The option \(1 - \frac{365 \times 364}{366 \times 366}\) is the complement: the chance that at least two of the three share a birthday. \(\frac{1}{366^{3}}\) is the chance that the three birthdays are three particular stated days, and \(\frac{1}{366 \times 365 \times 364}\) does not correspond to the event asked about; both are far too small.
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