Suppose that 12% of the women who purchase over-the-counter pregnancy testing ki

Question

Suppose that 12% of the women who purchase over-the-counter pregnancy testing kits are actually pregnant. For a particular brand of kit, if a woman is pregnant, the test will yield a positive result 96% of the time and a negative result 4% of the time. If she is not pregnant, the test will yield a positive result 5% of the time and a negative result 95% of the time. Suppose the test comes up positive. Let event A be Positive test. Let event B be Pregnant.

(1) P(A|B) =?

(2) P(A|B')=?

(3) What is the probability that she is really pregnant? (i.e. What is P(B|A) (the probability that she is really pregnant under the condition that the test comes up positive)?)

Explanation / Answer

Ans:

A=positive test

B=pregnant

1)P(positive/pregnant)=P(A/B)=0.96

P(negative/pregnant)=P(A'/B)=1-0.96=0.04

2)P(positive/not pregnant)=P(A/B')=0.05

P(negative/not pregnant)=(A'/B')=0.95

P(B)=0.12,P(B')=1-0.12=0.88

3)Bayes Theorm:

P(pregnant/positive)=P(B/A)=P(A/B)*P(B)/P(A)

Where,P(A)=P(A/B)*P(B)+P(A/B')*P(B')

P(B/A)=P(A/B)*P(B)/[P(A/B)*P(B)+P(A/B')*P(B')]

           =0.96*0.12/[0.96*0.12+0.05*0.88]

          =0.1152/0.1592=0.7236

Hence,P(B/A)=0.7236

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.