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
[tex]P(A) = \dfrac{4}{8} = \dfrac{1}{2}[/tex]
because 4 out of 8 numbers are prime: 2, 3, 5 and 7.
Similarly, we have
[tex]P(B) = \dfrac{4}{8} = \dfrac{1}{2}[/tex]
because 4 out of 8 numbers are odd: 1, 3, 5 and 7.
Finally,
[tex]P(A \land B) = \dfrac{3}{8}[/tex]
because 3 out of 8 numbers are prime and odd: 3, 5 and 7.
So, applying the formula, we have
[tex]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}[/tex]
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.
Using the probability concept, it is found that:
P(prime|odd) = 0.75.
A probability is given by the number of desired outcomes divided by the number of total outcomes.
The probability P(primelodd) is the probability of choosing a prime number, counting only odd numbers. We have that there are 4 odd numbers(1, 3, 5 and 7), of which 3 are prime(3, 5 and 7), hence:
P(prime|odd) = 3/4 = 0.75.
More can be learned about probabilities at https://brainly.com/question/14398287
