Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A class has students. Assume that each student's birthday is equally likely to be any of days, that the students are independent, and that leap days are ignored. Find the probability that all students have different birthdays. Find the probability that at least two share a birthday. Then find the smallest class size for which the probability of a shared birthday is more than . Give the probabilities to four places.
All different: for objects in equally likely boxes. For matching, counts the arrangements with no match, and exactly matches have probability .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. All different. The probability is .
Step 2. A shared birthday. The complement is .
Step 3. The smallest class. The chance of a shared birthday is about for students and about for students. So is the smallest class for which it is more than .
Answer: , and students.
Watch out. A class of does not need . The number of pairs, , is what makes a shared birthday so likely.
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