Question 1036516: i am really not sure how to go about this question ,i've tried it out using factorial but i still don't get it ,
How many people must be in a room so that with probability greater than 0.6, two of them have been born on the same day of the week?
Thanks
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
Obviously if there are 8 or more people or more
in the room the probability is 1.
We go for the complement event, the event that n people
all have different birth days. Then subtract from 1.
[Notice that "birth day" does not mean "birthday". LOL]
Case 1: One person in the room.
The probability is 0 that 2 have the same birth day, for
there aren't 2 people in the room.
Case 2: Two people in the room.
We can assign them all different birth days in 7P2 = 42 ways.
We can assign them birthdays in 7*7=49 ways. The
probability that they have different birth days is
42/49. So the probability that they have the same
birthday 1-42/49 = 1/7 = 0.142871429
Case 3: Three people in the room.
We can assign them all different birth days in 7P3 = 210 ways.
We can assign them birthdays in 7*7*7=343 ways. The
probability that they have different birth days is
210/343. So the probability that they have the same
birthday 1-210/343 = 19/49 = 0.387755102
Case 4: Four people in the room.
We can assign them all different birth days in 7P4 = 840 ways.
We can assign them birthdays in 7*7*7*7=2401 ways. The
probability that they have different birth days is
840/2401. So the probability that they have the same
birthday 1-840/2401 = 223/343 = 0.6501457726
That's the answer, 4 people, but let's follow through
for instructive purposes.
Case 5: Five people in the room.
We can assign them all different birth days in 7P5 = 2520 ways.
We can assign them birthdays in 7*7*7*7*7=16807 ways. The
probability that they have different birth days is
840/2401. So the probability that they have the same
birthday 1-2520/16807 = 2041/2401 = 0.850062474
Case 6: Six people in the room.
We can assign them all different birth days in 7P6 = 5040 ways.
We can assign them birthdays in 7*7*7*7*7*7=117649 ways. The
probability that they have different birth days is
5040/117649. So the probability that they have the same
birthday 1-5040/117649 = 112609/117649 = 0.9571607060
Case 7: Seven people in the room.
We can assign them all different birth days in 7P7 = 5040 ways.
We can assign them birthdays in 7*7*7*7*7*7*7=823543 ways. The
probability that they have different birth days is
5040/2401. So the probability that they have the same
birthday 1-5040/823543 = 116929/117649 = 0.993880101
Edwin
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