Using conditional probability, it is found that there is a 0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

<h3>What is Conditional Probability?</h3>

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, the events are:

Event A: Positive test result.

Event B: Pregnant.

The probability of a positive test result is composed by:

99% of 10%(truly pregnant).

2% of 90%(not pregnant).

Hence:

$P(A) = 0.99(0.1) + 0.02(0.9) = 0.117$

The probability of both a positive test result and pregnancy is:

$P(A \cap B) = 0.99(0.1)$

Hence, the conditional probability is:

$P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462$

0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

You can learn more about conditional probability at brainly.com/question/14398287

7 0

2 years ago

Which answer choice shows 3.002 written in expanded form?A) 3 + 0.2B) 3 + 0.02C) 3 + 0.002D) 3+ 0.0002

Elanso [62]

SOLUTION

We want to know which answer choice shows 3.002 written in expanded form

To do this let us subtract 3.002 from 3, we have

We got 0.002

So the expanded form is

$3+0.002$

Hence the correct answer is option C

8 0

1 year ago