Question

Home pregnancy tests, like all medical tests, are not perfectly accurate. Suppose one such test has an accuracy level such that if a woman is pregnant there is a 12% chance that the test will fail to show it. On the other hand, if the woman is not pregnant there is still a 3% chance that the test will incorrectly say that she is. Show all the steps including identification of what formulas/properties you used. Points will be deducted from answers if only the final answer is provided.

a) If Joan is pregnant what is the chance that the test will detect it?
b) Based on other factors, Joan estimates there is an 80% chance that she is pregnant.
Therefore, Joan takes a pregnancy test. What is the probability that the test will indicate that she is pregnant?
c) Joan looks at the test result and it does indicate that she is pregnant. Given this test result, what is the probability that she is pregnant?
d) Pamela has also taken this particular home pregnancy test. Before taking the test, she estimated there was a 75% chance that she

Answer 1

Joan has an 88% chance the test will detect her pregnancy if she is indeed pregnant. The probability the test will indicate she is pregnant is 0.71, and the probability that Joan is actually pregnant given a positive test result is approximately 99.15%.

Step-by-step explanation:

To answer the questions about the accuracy of home pregnancy tests and the probabilities involved with pregnant women being tested, we'll use concepts of conditional probability and Bayes' theorem.

a) Probability the test will detect Joan's pregnancy

If Joan is pregnant, the test has an 88% chance of detecting the pregnancy (100% - 12% chance of not detecting it).

b) Probability the test will indicate Joan is pregnant

Using the total probability theorem:

P(Test positive) = P(Test positive | Joan is pregnant)P(Joan is pregnant) + P(Test positive | Joan is not pregnant)P(Joan is not pregnant)

= 0.88 * 0.80 + 0.03 * 0.20

= 0.704 + 0.006

= 0.71

c) Probability Joan is pregnant, given a positive test result

We use Bayes' theorem here:

P(Joan is pregnant | Test positive) = P(Test positive | Joan is pregnant)P(Joan is pregnant) / P(Test positive)

= (0.88 * 0.80) / 0.71

= 0.704 / 0.71

≈0.9915 or about 99.15%

Answer 2

Answer:just go to the doctor

Step-by-step explanation: