An otherwise empty room contains 6x1026 air molecules How ma

An otherwise empty room contains 6x10^26 air molecules. How many times more probable is it that they are split exactly 50-50 between front and back halves of the room than that there is a 49-51 split?

Solution

The number of microststes with half on either side is ( since each molecule can be on either side) 2^6x10²⁶

out of which the probability of 49:51 is I the 1% on the rear half can be any of he 6x1026molecules)

hence odd of having 49:51 is 6x 10²⁶/  2^n(6x10²⁶)