A pregnancy test kit is 98.5% accurate for true positive result, i.e. the result is positive when the tester is actually pregnant. If she is not pregnant, however, it may yield a 0.8% false positive. Suppose a woman using this pregnancy kit is 60% at risk of being pregnant.

  1. Not sure about her first test which turned out to be negative, the woman decides to take the test again. This second test, however, turns out to be positive. Assuming the two test are independent, find the probability that she is actually pregnant Ans: Please Help

  2. Now she is so confused whether or not she is pregnant. So she take the tests n more times and the results for these n more tests are all positive. Find the minimum value for n so that she can be at least 99.99% sure of pregnancy, assuming all test are independent. Ans: Please Help

  • $\begingroup$ I found P(preg|+)P(+) and P(preg|-)P(-) = P(Preg) But some of my friend told me that it is P(Preg| - and +) so I confused how to solve this out $\endgroup$ – Wongsakorn Feb 29 '16 at 2:11
  • $\begingroup$ I tried the Geometric Prob. in 2. And I don't know what to do next. So I need to verify. Because I don't think both it is really correct $\endgroup$ – Wongsakorn Feb 29 '16 at 2:12
  • $\begingroup$ You need to use Bayes's Rule. en.wikipedia.org/wiki/Bayes%27_theorem . (Addendum:) and your friends are correct. $\endgroup$ – Christopher Carl Heckman Feb 29 '16 at 2:13

For part 1, use Bayes's Rule. (Here, I use $-+$ to mean the first test is negative and the first one positive.)

$$\eqalign{P(Preg \mid -+)&= {P(Preg \cap -+)\over P(-+)\mathstrut} \cr&= {P(Preg \cap -+)\over P(Preg \cap -+)+ P(\overline{Preg} \cap -+)} \cr&= {P(Preg) P( -+\mid Preg)\over\mathstrut P(Preg) P(-+\mid Preg)+ P(\overline{Preg}) P(-+ \mid \overline{Preg})} \cr&= {\mathstrut(0.6)(0.015)(0.985) \over (0.6)(0.015)(0.985)+(0.4)(0.992)(0.008)} \approx 0.74\cr} $$

  • $\begingroup$ Now, to find P(-+|Preg) you mean it is equals to P(-|Preg)*P(+|Preg) right ? $\endgroup$ – Wongsakorn Feb 29 '16 at 2:31
  • $\begingroup$ @Wongsakorn Yes. $\endgroup$ – Christopher Carl Heckman Feb 29 '16 at 6:06
  • $\begingroup$ @trueblueanil Probability tends to be non-intuitive, so there is no "reality check" here. If you leave out the 0.6 and 0.4 factors (like I did originally), you get about 65%. $\endgroup$ – Christopher Carl Heckman Feb 29 '16 at 6:08

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