Question

According to the general equation for conditional probability, if P(A, B) =4/7

and P(B) =7/8 - what is P(AIB)?

Answer 1

Answer:

$P(A|B) = \frac{32}{49}$

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

In which

P(A|B) is the probability of event A happening, given that B happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(B) is the probability of B happening.

We have that:

$P(A \cap B) = \frac{4}{7}, P(B) = \frac{7}{8}$

So

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{4}{7}}{\frac{7}{8}} = \frac{4}{7}*\frac{8}{7} = \frac{32}{49}$

Then

$P(A|B) = \frac{32}{49}$