![]() |
|
|
|||||||
| FlashChat | Actuarial Discussion | Preliminary Exams | CAS/SOA Exams | Cyberchat | Around the World | Suggestions |
Life Actuarial Jobs | Salary Surveys | DW Simpson & Co. | Casualty Jobs |
![]() |
|
|
Thread Tools | Display Modes |
|
#1
|
|||
|
|||
|
I can't seem to figure this one out. Can anyone help?
"8 people were all born in January. What is the probability that at least 2 of them have the same birthday?" I think the total # of ways 8 people can be born in January is 31 P 8, but that's as far as I can get. Thanks! |
|
#2
|
||||
|
||||
|
Approach it as 1 - Prob(no matches)
prob(no matches) = prob(#1 can have any b-day) x prob(#2 doesn't match #1)x(#3 doesn't match #2 or #1)x(#4 doesn't match #3 or #2 or #1) ... = (31/31)x(30/31)x(29/31)x(28/31)x...x(24/31) or (31 P 8)/(31^8) = .627 This problem shows up a lot in textbooks as...How many people need to be in a room so that there is a greater than 50% chance that there is at least one shared birthday. Need to find n such that (365 P n)/(365 ^ n) > .5 Turns out that n >= 23 !!!! |
|
#3
|
|||
|
|||
|
Use 1 - P(no one shares a Birthday)
This is the same as 1 - 31P8/31^8 |
|
#4
|
||||
|
||||
|
Two errors I made: (31 P 8)/(31 ^ 8) = .373, and 1 - .373 = .627 which is the answer to the question.
And my eights became emoticons somehow... |
|
#5
|
|||
|
|||
|
Thanks for your help!!
|
![]() |
| Thread Tools | |
| Display Modes | |
|
|