Explanation:
a probability is always the ratio of
desired cases / totally possible cases
I assume the meaning here is "of any year".
so, e.g. August of 1998 is the same for this experiment as August 2001.
similar for the day of the week.
we also assume that each day of the week has the same number of appearances per month. otherwise we would need to know the precise years of all 8 people to establish how many Mondays and how many Tuesdays ... are in each month in question.
so, for the totally possible cases we have
12 months
7 days of the week
so, in total
12 × 7 = 84 cases
so, this is like rolling a die with 84 sides.
the probabilty to have a birthday in a specific month and in a shortcut day of the week is like rolling that die :
1/8
the probability for 2 people to have the same "type" of birthday is like rolling the same number on that die 2 times in a row :
1/84 × 1/84 = 1/7056 = 0.000141723...
(remember, rolling 2 6s on a regular die is 1/6 × 1/6 = 1/36).
so, each pair of people we can create out of the 8 people has a probability of 1/7056 to have the same type of birthday.
how many pairs can we create ?
this is combinations of 2 outdoor out of 8, where the order (sequence of the pulled items) does not matter :
C(8, 2) = 8!/(2! × (8-2)!) = 8×7/2 = 4×7 = 28
that means we can create 28 different pairs of people out of the 8.
the probability that 2 of these 8 people have the same type of birthday is
28 × 1/7056 = 1/252 = 0.003968253...
now, the probability that at least 2 have the same birthday would be the sum of 2 people having the same birthday, 3 people having the same birthday, 4 people having the same birthday, ..., all 8 having the same birthday.
that can be a complex calculation.
we can also say the probability of at least 2 people having the same birthday is the complementary probability of all 8 people having different birthdays :
P(2+ the same birthday) =
= 1 - P(everybody different birthday)
to calculate the probabilty that all 8 people have different birthdays, let's think again about the totally possible cases :
each person has 84 possibilities.
so, 8 people have
84×84×84×84×84×84×84×84 = 84⁸
total possibilities.
about the desired cases :
the first person has 84 "choices".
the second person has then one choice less to be different to the first person : 83.
the third person has 2 options less to be different : 82.
4. : 81
5. : 80
6. : 79
7. : 78
8. : 77
so, we have
84×83×82×81×80×79×78×77 possible desired cases of different birthdays.
the probability for that is then
84×83×82×81×80×79×78×77/84⁸ =
= P(84, 8)/84⁸ = 0.709127565...
the probabilty for at least 2 people out of the 8 have the same birthday is then
1 - 0.709127565... = 0.290872435...