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Britney can be homozygous $HH$ or heterozygous $Hh$ with equal probability. Hemophilia is a mostly inherited genetic disorder. A test to detect a dominant allele $h$, responsible for the disorder, is carried out. The test has $85\%$ reliability in heterozygous women (with $Hh$ genotype), that is, it successfully detects the presence of the allele $h$ in $85\%$ of the cases, while in homozygous women (with $HH$ genotype) it fails to detect $h$ in $1\%$ of the cases. We want to calculate the following probabilities: $P (\text{Britney}\,Hh | \text{test was positive})$ and $P(\text{Britney}\,HH | \text{test was negative})$

I am not sure for the correct interpretation of the question, as I had to translate some terms I am not familiar with. With the little knowledge I have on statistics, I will make an attempt:

Prior probability Britney is homozygous or heterozygous $P(ΗΗ)= P(Hh) = 0.5$

$$P(E|Hh)= \text{Probability of a Positive Test Result given Britney is Heterozygous} = 0.85\\ \text{So, we have}\\ P(E|HH)= \text{Probability of a Positive Test Result given Britney is Homozygous} = 0.15$$

We want $$P(HH|E) = \text{Probability of Britney being Heterozygous given the test yields a Positive Result}$$

We also want $$P(Hh|E^c) = \text{Probability of Britney being Homozygous given the test yields a Negative Result}$$

So for a)

$$P(HH|E) = {P(E|HH) P(HH) \over P(E)} = {P(E|HH) P(HH) \over P(E|HH)P(HH) + P(E|{Hh}) P({Hh})}$$ and similarly for the second. Are these correct?

EDIT: Can you tell me if this is correct?

"$P(E|HH)= \text{Probability of a Positive Test Result given Britney is Homozygous} = 0.15$"

or is it "$P(E|HH)= \text{Probability of a Negative Test Result given Britney is Heterozygous} = 0.15$"?

Nhung Huyen
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    Please use proper TeX formatting when including texts within an equation from next time. – math-physicist Jun 01 '22 at 12:16
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    $P(Hh \mid E)$ is not the probability of Britney being Homozygous given the test yields a Negative Result but given the test yields a Positive Result. You want $P(Hh \mid E^c)$ – Henry Jun 01 '22 at 13:08
  • Your final expression looks reasonable if you exclude the possibility that Britney is $hh$. Female haemophiliacs are rare but not unknown – Henry Jun 01 '22 at 13:12
  • In a) you may have mixed homozygous / heterozygous or HH / Hh – Henry Jun 01 '22 at 14:31
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    Guys, any full solution? This one I don't understand! – Reyansh Laghari Jun 01 '22 at 14:32
  • @Henry is it OK now? – Nhung Huyen Jun 01 '22 at 15:31
  • Yes, I'd appreciate a full solution by someone knowledgeable!! Mine was just an attempt... – Nhung Huyen Jun 01 '22 at 16:04
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    You may still have mixed homozygous / heterozygous. I would have thought you wanted $P(Hh\mid E) = {P(E\mid Hh) P(Hh) \over P(E)} = {P(E\mid Hh) P(Hh) \over P(E\mid HH)P(HH) + P(E\mid {Hh}) P({Hh})}$ and $P(HH\mid E^c) = {P(E^c\mid HH) P(HH) \over P(E^c)} = {P(E^c\mid Hh) P(Hh) \over P(E^c\mid HH)P(HH) + P(E^c\mid {Hh}) P({Hh})}$ – Henry Jun 01 '22 at 16:30
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    Related, perhaps helpful: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Jun 02 '22 at 14:10

2 Answers2

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As I understand from your setting, you could approach the problem as follows:

Suppose you let $T$ be the event "the test is positive" (so the test detects the allele $h$), $HH$ the event "homozygous genotype", and $Hh$ the event "heterozygous genotype". We wish to compute the probabilities $Pr(Hh|T)$ and $Pr(HH|T^c)$.

We know the following:

  • $Pr(HH)=Pr(Hh)=1/2$, so the probability of having either genome is the same.
  • $Pr(T|Hh)=0.85$ from which you can deduce that $Pr(T^c|Hh)=0.15$.
  • $Pr(T^c|HH)=0.01$ from which you can deduce that $Pr(T|HH)=0.99$.

From there, using Bayes' theorem, the law of total probability and remembering that the probability of having either genome is equal, we can directly compute \begin{align} Pr(Hh|T)=\frac{Pr(T|Hh)Pr(Hh)}{Pr(T)} &=\frac{Pr(T|Hh)Pr(Hh)}{Pr(T|Hh)Pr(Hh)+Pr(T|HH)Pr(HH)}\\ &=\frac{0.85\cdot 1/2}{(0.85+0.99)/2}\approx 46.2\%, \end{align} and \begin{align} Pr(HH|T^c)=\frac{Pr(T^c|HH)Pr(HH)}{Pr(T^c)}&=\frac{Pr(T^c|HH)Pr(HH)}{Pr(T^c|HH)Pr(HH)+Pr(T^c|Hh)Pr(Hh)}\\ &=\frac{0.01\cdot 1/2}{(0.01+0.15)/2}=6.25\%. \end{align}

Edit: I've confused the original statement and so had wrong calculations. The procedure however stays the same.

seboll13
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  • I don't think this is the correct interpretation of "The test has $85%$ reliability in heterozygous women. I'd rather assume it means that in $15%$ of the cases it fails to detect the allele h, while it exists. – Nhung Huyen Jun 03 '22 at 16:46
  • @NhungHuyen I've updated my answer with the correct calculations. The statement seemed confusing at start so I mixed up two probabilities. The answer should be good now :) – seboll13 Jun 04 '22 at 09:59
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By the law of total probability, $$P(Positive)=P(Positive|HH)P(HH)+P(Positive|Hh)P(Hh)=0.5\cdot0.99+0.5\cdot0.85=0.5(0.99+0.85)=0.5\cdot1.84=0.92$$ and so P(Negative)=0.08 (can also be calculated from the law of total probability).
(a)By Bayes Theorem, $$P(Hh|Positive)=\frac{P(Hh\cap Positive)}{P(Positive)}$$ $$=\frac{0.85\cdot 0.5}{0.92}=\frac{85}{184}≈0.4619=46.19\% $$
Similarly, (b) By Bayes Theorem, $$P(HH|Negative)=\frac{P(HH\cap Negative)}{P(Negative)}$$ $$=\frac{0.01\cdot 0.5}{0.08}=\frac{1}{16}=0.0625=6.25\%$$

insipidintegrator
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