Find the probability of being same

[Diganta281]

This question is somewhat related to computer but deals with a big lot of probability.I have been given two files (A and B) of size 8192 bytes and is asked to find the probability of that file B is same as file A.I know that for two files to be same, they should have same bit sequence i.e same sequence of 1s and 0s.
As the files are of 8192 bytes so they have (8192*8) i.e 65536 bits.
Now our sample space (S) includes all the probable bit sequences of file A.
Using the formulae variation^length, I got that n(S)=2 ^65536
[Since , variations are 1 and 0 i.e 2 variations and length =65536 bits]
Now , Since file B can have only one sequence of bits
\therefore n(E)=1 [let E denote the set of favourable outcomes to the event that file B is same as file A]
\therefore probability of the matching of the two files = n(E)/ n(S) =1/2^65536 = 2^-65536

Upto this it is clear to me but the real problem starts in the next part :
In the next part of problem it is given that file A as well as file B have 16384 bits filled with 1s and the rest bits filled with 0s.
Now certainly the above found probability would increase as in the above part we have considered that both file A and B can have as many 1s and 0s required to fill the complete sequence of bits.
Now I am not able to figure out what the probability of both the files being same would be!
Plz help me soon...

 

[chisigma]

This question is somewhat related to computer but deals with a big lot of probability.I have been given two files (A and B) of size 8192 bytes and is asked to find the probability of that file B is same as file A.I know that for two files to be same, they should have same bit sequence i.e same sequence of 1s and 0s.
As the files are of 8192 bytes so they have (8192*8) i.e 65536 bits.
Now our sample space (S) includes all the probable bit sequences of file A.
Using the formulae variation^length, I got that n(S)=2 ^65536
[Since , variations are 1 and 0 i.e 2 variations and length =65536 bits]
Now , Since file B can have only one sequence of bits
\therefore n(E)=1 [let E denote the set of favourable outcomes to the event that file B is same as file A]
\therefore probability of the matching of the two files = n(E)/ n(S) =1/2^65536 = 2^-65536

Upto this it is clear to me but the real problem starts in the next part :
In the next part of problem it is given that file A as well as file B have 16384 bits filled with 1s and the rest bits filled with 0s.
Now certainly the above found probability would increase as in the above part we have considered that both file A and B can have as many 1s and 0s required to fill the complete sequence of bits.
Now I am not able to figure out what the probability of both the files being same would be!
Plz help me soon...

Your approach to the problem would be fair if any combination of 8192 bytes had the same probability, and this clearly is not true ... I think your question isn't easy to answer ...

Kind regards

$\chi$ $\sigma$