In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any two players with the same birthdays? (just assume 365 days a year and don't have to do the exact calendar month and day, use the day number from 1 to 365)

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.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3