(Assume the 7 days Sun, ..., Sat are all equally likely days to be born on.)
2007-11-08 13:49:03 · 3 answers · asked by Kent H 1 in Science & Mathematics ➔ Mathematics
It would be (7/7) * (6/7) * (5/7) * (4/7) * (3/7) * (2/7) = 6!/7^5. In other words, the first person can be born on any day, the second person can be born on any day except the day the first person was born on, the third person can be born on any day except the day the first two persons were born on, etc.
2007-11-08 13:59:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
there are 6 people:
_ _ _ _ _ _
probability of the first person being born on a day of the week is 1.
1*_ _ _ _ _
probability of second person being born on a different day is 6/7.
1*(6/7)*_ _ _ _
probability of third person being born on a different day is 5/7.
1*(6/7)*(5/7)*_ _ _
probability of fourth person being born on a different day is 4/7.
1*(6/7)*(5/7)*(4/7)*_ _
probability of fifth person being born on a different day is 3/7.
1*(6/7)*(5/7)*(4/7)*(3/7)* _
probability of 6th person being born on a different day is 2/7.
1*(6/7)*(5/7)*(4/7)*(3/7)* (2/7)
= 720/16807
=0.043
2007-11-08 22:02:06 · answer #2 · answered by James B 2 · 0⤊ 0⤋
7/7*6/7*5/7*4/7*3/7 *2/7= 7!/7^6 =0.0428
2007-11-08 22:00:20 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋