Question

A number is selected, at random, from the set {1,2,3,4,5,6,7,8}.

Answer 1

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

$P(A) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are prime: 2, 3, 5 and 7.

Similarly, we have

$P(B) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are odd: 1, 3, 5 and 7.

Finally,

$P(A \land B) = \dfrac{3}{8}$

because 3 out of 8 numbers are prime and odd: 3, 5 and 7.

So, applying the formula, we have

$P(\text{prime } | \text{ odd}) = \dfrac{P(\text{prime and odd})}{P(\text{odd})} = \dfrac{\frac{3}{8}}{\frac{1}{2}} = \dfrac{3}{8}\cdot 2 = \dfrac{3}{4}$

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.