2006-06-15 01:55:07 · 6 answers · asked by iambored_2287 1 in Science & Mathematics ➔ Mathematics
Suppose the people are p1, p2, p3, p4, p5, p6, p7.
The probability that p1 was born on Monday and nobody else was is:
1/7 * (6/7)^6 where 1/7 is the probability a person is born on Monday and 6/7 is the probability someone isn't born on Monday.
The same is true for the probability that p2 is born on Monday and nobody else is, etc.
So the total is
7 * (1/7)*(6/7)^6
= (6/7)^6
= 46656/117649
= .3965694565
2006-06-15 02:03:46 · answer #1 · answered by fatal_flaw_death 3 · 0⤊ 0⤋
The probability that a person wil be born on monday is 1/7, and the probability they won't is 6/7.
For each person the probability that they will be born on
monday and the rest aren't is: 1/7*(6/7)^6 .
But there are 7 people so we have to multiply this by 7
and so the probability of exactly one being born on
monday = (6/7)^6 = 0.39656...or about 39.7%
2006-06-15 09:17:40 · answer #2 · answered by Jimbo 5 · 0⤊ 0⤋
Lets assume the seven people are chosen randomly. You are not at a conference of twins. You are not choosing babies in a hospital etc.
Then for any particular person the chance of them being not born on a monday is 6/7 while the chance they are born on a Monday is 1/7.
Lets call the people A,B,C,D,E,F,G.
The odds A is born on monday and the rest are not. is
(1/7) * (( 6/7)^6)
This is also the odds for B being born on monday and the rest not.
Same as for C,D,E,F, and G.
So the odds for One of Seven persons being born on monday and the rest not are.
7 * ( 1/7) * ( (6/7)^6)
Or simplifying just (6/7)^6
This is 46656 / 117649
or approximately 0.39656945660396603456
2006-06-15 09:18:15 · answer #3 · answered by Anonymous · 0⤊ 0⤋
100%
2006-06-15 09:00:40 · answer #4 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
(1/7)*(6/7)^6
2006-06-15 09:18:11 · answer #5 · answered by sh.akbari 2 · 0⤊ 0⤋
7x1/7(6/7)^6
2006-06-15 10:56:24 · answer #6 · answered by gari 3 · 0⤊ 0⤋