The Birthday Problem. The Problem In a group of 50 students, what is the probability that at least two students share the same birthday?

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Presentation transcript:

The Birthday Problem

The Problem In a group of 50 students, what is the probability that at least two students share the same birthday?

Assumptions Only 365 days each year. Birthdays are evenly distributed throughout the year, so that each day of the year has an equal chance of being someone’s birthday. Note: The accuracy of the probability calculations that we do here depends on whether or not these assumptions are valid.

Take group of 5 people…. Assuming independence and using classical approach: P(A) = 365/365 × 364/365 × 363/365 × 362/365 × 361/365 = So, then P(A C ) = = Let A = event no one in group shares same birthday. Then A C = event at least 2 people share same birthday. That is, about a 3% chance that in a group of 5 people at least two people share the same birthday.

Take group of 23 people…. Assuming independence and using classical approach: P(A) = 365/365 × 364/365 × … × 343/365 = So, then P(A C ) = = Let A = event no one in group shares same birthday. Then A C = event at least 2 people share same birthday. That is, about a 51% chance that in a group of 23 people at least two people share the same birthday.

Take group of 50 people…. Assuming independence and using classical approach: P(A) = 365/365 × 364/365 × … × 316/365 = 0.03 P(A C ) = = 0.97 Let A = event no one in group shares same birthday. Then A C = event at least 2 people share same birthday. That is, about a 97% chance that in a group of 50 people at least two people share the same birthday.