Chemistry nmr spectrum question

These are mass spectra, not NMR spectra.

Since 79Br and 81Br each make up 50% of all bromine atoms:

Calculate the probability that a Br2 contains:

(a) 2 x 79Br (m/z = 158)
(b) 1 x 79Br + 1 x 81Br (m/z = 160)
(c) 2 x 81Br (m/z = 162)

Now compare the ratio of these probabilities.

Sorry for the wrong term
Wouldn’t the probability for all just be 1/4 since they both isotopes have a probability of 1/2?

1/4 for 2 x 79Br and 2 x 81Br, but not for 1 x 79Br + 1 x 81Br.

Think about it. Do those probabilities add up to 1?

Ohh, it should be 1/2 because Br 79 and Br 81 can occur in two ways so 1/4*2=1/2?
Thank you

Correct.

Now that means the ratio of p(m/z = 158) to p(m/z = 160) to p(m/z = 162) is 1/4 to 1/2 to 1/4

Try scaling up that ratio and see which option matches it best