Hi I have a dobt in question Suppose the probability of an individual being born on any particular day of the year is given by 1/365. Suppose now that a group has 3 individuals. What is the probability that at least two of these individuals will share a birthday? Here the solution displays Let all 3 persons do not share the same birthday whose probability based on above argument is 1/365 *364/365 * 363/365*365 = 365*364*363/3653 Thus, probability of at least 2 persons having the same birthday is 1-365*364*363/3653 = 0.008204 My doubt : Isn't it the probability of all three sharing the birthday. we haven't taken into account the probability of two persons sharing the same b'day. As P(at least 2)= P(X=2)+P(X=3). So shouldn't we calcualte P(X=2) and then add into it?
We've to calculate P(at least 2 ppl. share b'day) = P(2 share) + P(all 3 share), & P(no one share) + P(2 share) + P(all 3 share) = 1 So, they've calculated [ 1- P(no one share) ] which is the required probability. Clear??
