Question

this is a problem calculating probability. consider some physical system which is equally likely to be in one of any of 1000 unique microstates. let's say the system is in a macrostate with physical property 'a' for 10 of these microstates. let's say the system is in a macrostate with physical property 'b' for 5 of these microstates. what is the relative probability (i.e., how many times more or less likely) that the system has physical property 'a'?

Answer 1

The system is two times more likely to exhibit physical property 'a' compared to property 'b' because property 'a' has twice as many microstates associated with it.

Step-by-step explanation:

The question asks how many times more or less likely it is for a physical system to have physical property 'a' as compared to property 'b'.

To calculate this, we can use the probability of each macrostate, which is directly proportional to the number of microstates it encompasses.

Since there are 10 microstates for property 'a' and 5 for property 'b', the probability for 'a' is P(a) = 10/1000 = 0.01, and for 'b' it's P(b) = 5/1000 = 0.005.

The relative probability of the system having the property 'a' rather than 'b' is therefore P(a) / P(b) = 0.01 / 0.005 = 2.

This means that the system is two times more likely to exhibit property 'a' compared to property 'b'.