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1 vote
1 answer
97 views

Why isn't adding the ways to achieve every mutually exclusive outcome giving me the denominator in the birthday problem for four people?

Why isn't adding the ways to achieve every mutually exclusive outcome giving me the denominator in the birthday problem for four people? $$\binom{4}{2} \cdot 365 \cdot 364 +\binom{4}{3} \cdot 365 \...
user avatar
2 votes
2 answers
137 views

How do the graphs of the "Birthday Probability" functions work?

The question was to find the probabilities that: At least two people have the same birthday Only two people have the same birthday Nobody has same birthdays in a room of n people. I did find the ...
AltercatingCurrent's user avatar
2 votes
1 answer
1k views

Poisson paradigm in "near-birthday problem" example

I am presented with an example called the "near-birthday problem": What if we want to find the number of people required in order to have a 50-50 chance that two people would have birthdays within ...
Dom Fomello's user avatar
4 votes
4 answers
1k views

Birthday problem: using $^nC_r$.

In birthday problem say total number of people n < 365, then probability of all person having distinct birthday is given by, $$\frac{\text{total no. of ways of selecting $n$ numbers from $365$ ...
q126y's user avatar
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