\( P( \) no 2 people out of 5 share same What is the probability that in a class of 60 students, at least one pair of people share the same birthday?
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To determine the probability that at least one pair of students in a class of 60 shares the same birthday, we can use the complement rule. First, we calculate the probability that no students share a birthday. Assuming there are 365 days in a year and that each birthday is equally likely, the probability that the first student has a unique birthday is 1 (365/365). For the second student, the probability is 364/365, for the third student it’s 363/365, and so on, until the 60th student, who has 306/365 chances of getting a unique birthday. Therefore, the total probability for no shared birthdays is: \[ P(\text{no shared}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \ldots \times \frac{306}{365} \] Calculating this gives a very small value. Finally, to find the probability of at least one shared birthday, we find: \[ P(\text{at least one shared}) = 1 - P(\text{no shared}) \] This will yield a probability of around 99.99%, showing it’s highly likely that at least one pair shares a birthday. Understanding this concept can be a brain twister! It’s amazing how intuition regarding independent events can throw us off. In smaller groups, like just 5 people, the odds seem low, but scale it up to 60 and surprise! This probability often stuns those who think birthdays are spaced out evenly. The phenomenon behind it is linked to the pigeonhole principle—more “pigeons” than “holes” in our birthday scenario!
