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IAI Nov 10 Q3

D

dextar

Member
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??

 
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