prob doubt

[Akshay Jain]

wat is the prob that atleast 2 out of n ppl have the same bday...assume 365 days in a year nd all days r equally likely

[laracroft]

could it be like

prob. that atleast 2  have same bday is = { 1  -  365 C n / (365)^ n }

[Sinistral]

EDITED

              365Cn*n!
  1  -    ---------------
            (n+364)C364

[laracroft]

oh yeah.... i got my mistake... thanx sinistral !!

[laracroft]

hey sinistral,

why this denominator and why not n^365 ??

[Sinistral]

because n^365 implies more than 1 person can have more than 1 b'days.
(n+365-1)C(365-1) is equivalent to putting n balls in 365 bags such that any bag can have any no. of balls.

I hope I didn't miss anything this time .

[Akshay Jain]

@sinistral....pls explain d steps involved

[Akshay Jain]

1 more problem....if 6n tickts numbered 0,1,2,3,6n-1 are in bag nd 3 of dm are selectd find the prob that the sum of d no. on selectd tickets is 6n

[laracroft]

for the same reason i think the denominator shud be.... 365^n
because.... 1st person has 365 choices.
                  2nd.................365............
                  3rd..................365...........



                  n th person has 365 choices...

so multiplying them all gives us 365^n ... and it doesnt imply that any person can have more than one bday..

thats wot i thot

kindly verify !!

[Sinistral]

akshay jain wrote

1 more problem....if 6n tickts numbered 0,1,2,3,6n-1 are in bag nd 3 of dm are selectd find the prob that the sum of d no. on selectd tickets is 6n

(3+6n-1)C6n

what is the answer?

[Mauli]

i second you croft!

[Xiang]

it should be 365^n only :/ the chain goes like 364/365*363/365.....lalala

[Akshay Jain]

ans to 2nd ques is 3n/(6n-1)(6n-2)

[Sinistral]

I am confused between 365^n and n+365-1Cn.
the latter is: the number of 365 tuples such that their sum is n (each time) and the each member of 365 tuples can take zero or positive integer.

what extra things are being counted in the former??

[Sinistral]

I got my mistake but I do not know how to do it right, yet.

[Akshay Jain]

Ans to 1st ques is 1-(1-1/365)(1-2/165)(1-3/365)....(1-(n-1)/365)

[Sinistral]

hmmm..
means denominator will be 365^n. seems logical too. but i still cant think of what i am missing when I write n+365-1Cn

[Rizwan]

 prob that atleast 2 out of n ppl have the same bday will be as follows:
since each member have 365 days to chose from as their b day now we get total outcome as 365^n

now each person among n having different bday is 365!/(365-k)

so required prob= 1- (365!/(365-n)!365^n)

[Rizwan]

 CORRECTED
prob that atleast 2 out of n ppl have the same bday will be as follows:
since each member have 365 days to chose from as their b day now we get total outcome as 365^n
now each person among n having different bday is 365!/(365-n)!
so required prob= 1- (365!/(365-n)!365^n)

[Akshay Jain]

In an examination dere r 12 ques vd 4 choices each...if a person choses 1 of the 4 options at random dn find d prob of getting exactly 3 correct ans given that he attempts all d ques..