Question

Given that events A and B are independent with P ( A ) = 0.26 P(A)=0.26 and P ( B ∣ A ) = 0.65 P(B∣A)=0.65, determine the value of P ( B ) P(B), rounding to the nearest thousandth, if necessary.

Answer 1

Answer:

P(B) = 0.65

Step-by-step explanation:

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

$P(A) = 0.26, P(B|A) = 0.65$

They are independent events, which means that $P(A \cap B) = P(A)*P(B)$. So

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

$0.65 = \frac{P(A)*P(B)}{P(A)}$

$P(B) = \frac{0.65P(A)}{P(A)}$

$P(B) = 0.65$

Answer 2

Answer:

P(B) = 0.65

Step-by-step explanation:

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

In which

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

is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

They are independent events, which means that . So

Step-by-step explanation: