Probability of 5 Spin 1/2 Particles in No Magnetic Field

[moknight]

Homework Statement

Consider an ideal system of 5 non-interacting spin 1/2 particles in the absence of an external magnetic field. What is the probability that n of the five spins have spin up for each of the cases n = 0, 1, 2, 3, 4, 5?

Homework Equations

I'm guessing $\frac{N!}{n!(N-n)!}$

The Attempt at a Solution

I've done total number of arrangements Ω = $\frac{N!}{n!(N-n)!}$

done this for each case, n=0,1,2 etc.

Then p = $\frac{1}{Ω}$ for the corresponding Ω to get the probability.

This is a guess and I'm not really sure if I'm going the right way.

I'm getting probability-like values, such as p=0.2 for cases n=1 and n=4.

p=0.1 for cases n=2, n=3.

Would appreciate some help.

Thanks.

 

[vela]

Homework Statement

Consider an ideal system of 5 non-interacting spin 1/2 particles in the absence of an external magnetic field. What is the probability that n of the five spins have spin up for each of the cases n = 0, 1, 2, 3, 4, 5?

Homework Equations

I'm guessing $\frac{N!}{n!(N-n)!}$

The Attempt at a Solution

I've done total number of arrangements Ω = $\frac{N!}{n!(N-n)!}$

This isn't the total number of arrangements. You have 5 particles, each of which can be in one of two state. How many arrangements can you make? This is just like asking how many outcomes you can get from flipping 5 coins.

You also need to figure out what $\Omega$ represents if it's not the total number of arrangements.

You may want to read about the binomial distribution again.

done this for each case, n=0,1,2 etc.

Then p = $\frac{1}{Ω}$ for the corresponding Ω to get the probability.

This is a guess and I'm not really sure if I'm going the right way.

I'm getting probability-like values, such as p=0.2 for cases n=1 and n=4.

p=0.1 for cases n=2, n=3.

Would appreciate some help.

Thanks.