Applying the formula, you have:
A = the number is prime
B = the number is odd
I assume that with "random" you imply that all numbers can be chosen with the same probability. So, we have
![P(A) = \dfrac{4}{8} = \dfrac{1}{2}](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cdfrac%7B4%7D%7B8%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D)
because 4 out of 8 numbers are prime: 2, 3, 5 and 7.
Similarly, we have
![P(B) = \dfrac{4}{8} = \dfrac{1}{2}](https://tex.z-dn.net/?f=P%28B%29%20%3D%20%5Cdfrac%7B4%7D%7B8%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D)
because 4 out of 8 numbers are odd: 1, 3, 5 and 7.
Finally,
![P(A \land B) = \dfrac{3}{8}](https://tex.z-dn.net/?f=P%28A%20%5Cland%20B%29%20%3D%20%5Cdfrac%7B3%7D%7B8%7D)
because 3 out of 8 numbers are prime and odd: 3, 5 and 7.
So, applying the formula, we have
![P(\text{prime } | \text{ odd}) = \dfrac{P(\text{prime and odd})}{P(\text{odd})} = \dfrac{\frac{3}{8}}{\frac{1}{2}} = \dfrac{3}{8}\cdot 2 = \dfrac{3}{4}](https://tex.z-dn.net/?f=P%28%5Ctext%7Bprime%20%7D%20%7C%20%5Ctext%7B%20odd%7D%29%20%3D%20%5Cdfrac%7BP%28%5Ctext%7Bprime%20and%20odd%7D%29%7D%7BP%28%5Ctext%7Bodd%7D%29%7D%20%3D%20%5Cdfrac%7B%5Cfrac%7B3%7D%7B8%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cdfrac%7B3%7D%7B8%7D%5Ccdot%202%20%3D%20%5Cdfrac%7B3%7D%7B4%7D)
Note:
I think that it is important to have a clear understanding of what's happening from a conceptual point of you: conditional probability simply changes the space you're working with: you are not asking "what is the probability that a random number, taken from 1 to 8, is prime?"
Rather, you are adding a bit of information, because you are asking "what is the probability that a random number, taken from 1 to 8, is prime, knowing that it's odd?"
So, we're not working anymore with the space {1,2,3,4,5,6,7,8}, but rather with {1,3,5,7} (we already know that our number is odd).
Out of these 4 odd numbers, 3 are primes. This is why the probability of picking a prime number among the odd numbers in {1,2,3,4,5,6,7,8} is 3/4: they are literally 3 out of 4.