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osquare | 05:42 Mon 27th Sep 2021 | Business & Finance
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a store-bought pregnancy test produces a correct result 99% of the time when a woman is actually pregnant but is only 97% accurate when a woman is not pregnant. Assuming a woman has a 40% chance of actually being pregnant when she takes the test, and that the result comes out positive, what is the probability that she is actually pregnant


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If the result is positive then it is 99 %.
Id say its more tricky . My non maths but common sense answers 40 x0.99 + 60 x 0.03 whatever that gives.
Maybe around
Ignore me. Di'nt see the positive result bit
A long time since I've tried to tackle probabilities but I'd tackle it logically:
Assume 1000 women take the test.
40% of women are pregnant that take the test, so we have 600 not pregnant and 400 pregnant.
The test on the 600 non-pregnant women will give 582 negative and 18 positive results (97% accurate).
The test on the 400 pregnant women will give 396 positive and 4 negative results (99% accurate).
We have a total of 414 positive results and 396 of these are true positives.
So if a woman receives a positive result, the chance that she's actually pregnant is:
(396 / 414) x 100 = 95.65%

This is just my logical approach, so I could be wrong??
Am still not sure I agree with jackdaws result as we after take account of it might be a false positive which occurs in 3% of cases when woman isnt actually pregnant.
95.65% pregnat
Cant do maths but can do spreadsheets and I got

0.956521739 preggers

0.043478261 not pregnant
//// 0.956521739 preggers

0.043478261 not pregnant ////

..... which agrees with my answer above :)
Yes we agree. I just took to long to type it thinking no body else was interested
I did 0.99 x0.4
divided by 0.99x4+0.6*0.03
for pregnant.
Your way is much easier to follow and it would of been easier if I done it that way

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