Expand 4y(y-3)
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We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if
where is the whole sample space.
This implies that
So, let's compute the probability that all 22 footballer were born on different days.
The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day .
The second footballer can be born on any other day, so he has 364 possible birthdays:
the probability for the first two footballers to be born on two different days is thus
Similarly, the third footballer can be born on any day, except and
:
so, the probability for the first three footballers to be born on three different days is
And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.
The probability of all 22 footballers being born on 22 different days is thus
So, the probability that at least two footballers are born on the same day is
since the two events are complementary.