Respuesta :
Answer:
The correct option is (c). 0.92
Step-by-step explanation:
Let X = a woman is pregnant and Y = a pregnancy test is positive.
Given:
[tex]\\ P(X)=0.055\\P (Y|P)=0.99\\P(Y^{c}|X^{c})=0.995[/tex]
According to the Bayes' theorem, the conditional probability of an event A given than another event B has already occurred is:
[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}[/tex]
Use the above formula to compute the probability that a woman is pregnant given that the pregnancy test is accurate as follows:
[tex]P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})} \\=\frac{(0.99\times0.055)}{(0.99\times0.055)+((1-0.995)\times(1-0.055))}\\=0.9202\\\approx0.92[/tex]
Thus, the probability that a woman is pregnant given that the pregnancy test is accurate is 0.92.
The correct option is (c).
Answer:
Option c) 0.92
Step-by-step explanation:
We are given that at any given time about 5.5% of women (age 15-45) are pregnant;
Let Probability that women is pregnant, P(A) = 0.055
Probability that women is not pregnant, P(A') = 1 - P(A) = 1 - 0.055 = 0.945
Also, Let B be the event that test is positive;
Probability that pregnancy test is positive given that the woman is actually pregnant, P(B/A) = 0.99
Probability that pregnancy test is positive given that the woman is not actually pregnant, P(B/A') = 1 - 0.995 = 0.005
Now, we have to find that if the test yields a positive result what is the posterior probability that the woman is pregnant i.e.; P(A/B)
Using Bayes' Theorem we get;
P(A/B) = [tex]\frac{P(A) * P(B/A)}{P(A) * P(B/A)+ P(A') *P(B/A')}[/tex] = [tex]\frac{0.055*0.99}{0.055*0.99 + 0.945*0.005}[/tex] = 0.92
Therefore, the posterior probability of the hypothesis that the woman is pregnant is 0.92 .
