Question

Before going to the market, a pregnancy test has been applied to a population of 10,000 women from which 300 were actually pregnant. Suppose this new product  returns a “yes” result 99.5 % of the cases in which the person was actually pregnant and 98% of the cases returns  a “no” result when the person was not pregnant.

Now that the test is in the market: what is the probability that a random user (a women) of the test is not pregnant if the test returns a “yes “ result? Draw  the appropriated probability tree

Answer 1

P[ Women is pregnant ] = 300/10000

P[ Women is pregnant ] = 0.03

P[ Women is not pregnant ] = 1 - P[ Women is pregnant ]

P[ Women is not pregnant ] = 1 - 0.03

P[ Women is not pregnant ] = 0.97

P[ new product  returns a “yes” | she is actually pregant ] = 99.5% = 0.995

P[ new product  returns a “no” | she is not actually pregnant ] = 98% 0.98

P[ new product  returns a “yes” | she is not actually pregnant ] = 1 - P[ new product  returns a “no” | she is not actually pregnant ]

P[ new product  returns a “yes” | she is not actually pregnant ] = 1 - 0.98

P[ new product  returns a “yes” | she is not actually pregnant ] = 0.02

P[ new product  returns a “yes” ] = P[ new product  returns a “yes” | she is not actually pregnant ]*P[ Women is not pregnant ] + P[ new product  returns a “yes” | she is actually pregant ]*P[ Women is pregnant ]

P[ new product  returns a “yes” ] = 0.02*0.97 + 0.995*0.03

P[ new product  returns a “yes” ] = 0.0194 + 0.02985

P[ new product  returns a “yes” ] = 0.04925

We need to find P[ she is not actually pregnant | new product  returns a “yes” ]

P[ she is not actually pregnant | new product  returns a “yes” ] = P[ new product  returns a “yes” | she is not actually pregnant ]*P[ Women is not pregnant ] / P[ new product  returns a “yes” ]

P[ she is not actually pregnant | new product  returns a “yes” ] = 0.02*0.97 / 0.04925

P[ she is not actually pregnant | new product  returns a “yes” ] = 0.0194/0.04925

P[ she is not actually pregnant | new product  returns a “yes” ] = 0.3939