91Ó°ÊÓ

The complementary probability is the probability of an event not happening, which is \(1\) minus the probability of the event happening. This is useful when calculating the event we are interested in is complicated, but the complementary event is easier to solve. In this case, we want to find the probability of at least two people sharing a birth month. It is easier to find the complementary probability, which is the probability of no people sharing a birth month, and then subtract it from \(1\) to find the required probability.

Let there be \(n\) people in a room, each having a birth month randomly and independently. We will first calculate the probability that none of them have the same birth month. There are \(12^n\) possible ways their birth months can be arranged, considering all months to be equally likely. Out of these, there are \(12\) ways to choose the birth month for the first person, \(11\) ways for the second person (as their birth month can't be the same as the first person's), \(10\) ways for the third person, and so on, until we have only 1 way for the \(n^\text{th}\) person if \(n \leq 12\). So, the number of valid arrangements without a shared birth month is \(12 \times 11 \times 10 \times \cdots \times (12 - n + 1)\). The probability of no shared birth months can therefore be calculated as: $$ P(\text{no shared birth months}) = \frac{12 \times 11 \times 10 \times \cdots \times (12 - n + 1)}{12^n} $$

Now, using the complementary probability, we can find the probability of at least one shared birth month: $$ P(\text{at least one shared birth month}) = 1 - P(\text{no shared birth months}) = 1 - \frac{12 \times 11 \times 10 \times \cdots \times (12 - n + 1)}{12^n} $$ We want to find the smallest value of \(n\) such that: $$ P(\text{at least one shared birth month}) \geq \frac{1}{2} $$ To do this, we can calculate the probabilities for increasing values of \(n\) until the condition is satisfied. Note that for this problem, we need \(1 \leq n \leq 12\), since the probability of no shared birth month becomes zero if there are more people than there are months. Now let's calculate the probabilities for different values of \(n\): - For \(n=1\): \(P(\text{at least one shared birth month}) = 1 - \frac{12}{12} = 0 \leq \frac{1}{2}\) - For \(n=2\): \(P(\text{at least one shared birth month}) = 1 - \frac{12 \times 11}{12^2} \approx 0.08 \leq \frac{1}{2}\) - For \(n=3\): \(P(\text{at least one shared birth month}) = 1 - \frac{12 \times 11 \times 10}{12^3} \approx 0.23 \leq \frac{1}{2}\) - For \(n=4\): \(P(\text{at least one shared birth month}) = 1 - \frac{12 \times 11 \times 10 \times 9}{12^4} \approx 0.43 \leq \frac{1}{2}\) - For \(n=5\): \(P(\text{at least one shared birth month}) = 1 - \frac{12 \times 11 \times 10 \times 9 \times 8}{12^5} \approx 0.63 \geq \frac{1}{2}\) So, the smallest \(n\) that satisfies the condition is \(n=5\). Therefore, there need to be at least 5 people in a room for the probability of at least two people sharing a birth month to be greater than or equal to \(\frac{1}{2}\).