Question

If X and Y are mutually exclusive events with P(X) = 0.295, P(Y) = 0.32, then P(X⏐Y) =

Answer 1

Using conditional probability, it is found that P(X⏐Y) = 0.

What is Conditional Probability?

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

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

In which:

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

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

• P(A) is the probability of A happening.

In this problem, we have that X and Y are mutually exclusive events, hence $P(X \cap Y) = 0$, hence:

$P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.32} = 0$

More can be learned about conditional probability at brainly.com/question/14398287