Suppose a planet has 13 months in a year What is the probabi
Solution
I will explain the first question.
Well, the first person has 4 comparisons to make, but the second person was already compared to the first person, so there are only 3 comparisons to make. The third person then has 2 comparisons, the fourth person has 1. If you add up all possible comparisons (4 + 3 + 2 + 1) the sum is 10 comparisons, or combinations.
We\'ll suppose that the year has 30*13=390 days
The probability that a person does not have the same birthday as another person is 12 divided by 13 because there are 12 monts that are not a person\'s birthday. This means that any two people have a 12/13, or 92.3077 percent, chance of not matching birthdays.
If you multiply calculate (12/13)10, you\'ll find there\'s a 44.91 percent chance that all 10 comparisons contain no matches.
Consequently, the odds that there is a birthday match in those 10 comparisons is 1 – 0.4491 percent = 0.5509=55.09 percent
If we do the same with 13 people, we have:
(12+11+...+1)=78
The probability that at least two of them will have a birthday in same month is: 1-(12/13)28=0.9981=99.81 percent
For 15 person the probability that at least two of them will have a birthday in same month is 100%
If we do the same with 8 people, we have:
(7+6+5+...+1)=28
The probability that at least two of them will have a birthday in same month is: 1-(12/13)28=0.8937=89.37 percent

