A fetal test of genetic material is being developed to deter

A fetal test of genetic material is being developed to determine the eye color of a baby. The percentage of humans having brown eyes in given in D1, and the percentage having blue eyes is given in D2. Of those with brown eyes, the percentage carrying a certain genetic marker is given in D3, while of those with blue eyes, the percentage carrying the same marker is given in D4. If a person carries the marker, what is the probability that they will have blue eyes? (Enter your value in cell D6.)

Now, use these probabilities to design a simulation so that column A simulates one thousand people sampled at random for eye color and the marker. That is, enter one of “bm” (for brown without marker), “bM” (for brown with marker), “Bm” (for blue without marker), “BM” (for blue with marker) in each of A1:A1000. Check the validity of your model by giving in G1 the percentage of the simulation with “b”, in G2 the percentage of the simulation with “B”, in G3 the percentage of “b” the simulation with “M”, in G4 the percentage of “B” with “M”. Finally, in G6 give the proportion of “M” with “B”.

Solution

A fetal test of genetic material is being developed to determine the eye color of a baby.

Let, Br be the humans having brown eyes.

And we are given that,

P( Br) = 82% = 0.82

Let, Bl be the humans having blue eyes.

P(Bl) = 18% = 0.18

Let person carries the marker is C.

And percentage of marker for brown eyed is denoted by P(C / Br) = 10% = 0.1

similarly, percentage of marker for blue eyed is denoted by P(C / Bl) = 89% = 0.89

  If a person carries the marker, what is the probability that they will have blue eyes?

Here we use Baye\'s theorem.

We have to calculate P(Bl / C) = P(C / Bl) * P(Bl) / [P(C / Bl) * P(Bl) + P(C / Br) * P(Br) ]

P(Bl / C) = 0.89 * 0.18 / [ 0.89 * 0.18 + 0.10 * 0.82 ]

= 0.1602 / 0.24

P(Bl / C) = 0.6614