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So I am asking the following question for a friend of mine that studies medicine, I am myself a physics student so I have no formal training in probability theory but I hope I can formalize the question properly.


Problem

The question is: two parents both have a gene A and gene a, giving them total gene Aa or aA. When you either have aa or AA, you are defined as ill. Thus, assuming the gene is randomly passed on to a child, the chance of having an ill child is $\frac{1}{2}$. Now two children have been tested and had positive results, what is the chance two other children test negative for this illness?

This looks a lot like the boy girl paradox, where you see that a mother has a daughter and the probability of her having $2$ daughters is $\frac{1}{3}$.

They have been taught the Bayes' Theorem as follows: $$ P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\neg A)P(\neg A)} $$ Now I will explain my approach to the problem.

Interpretation

These parents have $4$ children, of which $2$ random have been tested, which resulted in a positive result, they are both ill. This is B, the constraint. Now we want to know the chance that the other two children do test negative.

Attempted solution

First we want to know what the chance of A is without constraint B. This can be interpreted as "The chance that two children out of the 4 are healthy". This should be $P(A)=\binom 4 2(\frac{1}{2})^2(\frac{1}{2})^2=\frac{6}{16}$. For the complement chance we use $P(A)+P(\neg A)=1 $ and thus $ P(\neg A)=1-P(A)=\frac{10}{16}$ Now for the chance P(B|A) I get lost, I have calculated and thought of all sorts of nonsense but I cannot get the right answer, which according to the teacher, should be $\frac{36}{64}$

It might be that I am tackling this problem from a completely wrong angle but this was the angle that I thought was best. Thanks in advance for reading and answering my question.

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  • $\begingroup$ This is not clear. What is a "positive" result? If the two "other" children are disjoint from the first two then surely the probabilities are independent. $\endgroup$ – lulu Jan 7 '18 at 14:30
  • $\begingroup$ @Lulu Positive means that the child has the disease and is ill. Negative means that the child is not ill. $\endgroup$ – Gijsv Jan 7 '18 at 15:28
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We'll approach this using sequences of length $4$, as there are four children. For each element in the sequence there are four possibilities (one of AA, Aa, aA, aa), so this gives $4^4=256$ possible sequences. Let's count all of them.

Sequences with

\begin{align} \text{exactly $0$ AA or aa} &\qquad 2^4 = 16 \\ \text{exactly $1$ AA or aa} &\qquad 2 \cdot 4 \cdot 2^3 = 64 \quad\text{(there are $4\cdot 2^3$ possibilities for AA)} \\ \text{exactly $2$ AA or aa} &\qquad \binom 4 2 2^4 = 96 \\ \text{exactly $3$ AA or aa} &\qquad 64 \quad \text{(note that this is the same to exacly $1$ Aa or aA)} \\ \text{exactly $4$ AA or aa} &\qquad 16 \end{align}

To check if we have counted all possible sequences, add all these numbers toghether. This gives $16+64+96+64+16=256$, so we have correctly counted all the different sequences.

Now, to answer your question. The information we have is that our sequence has at least $2$ AA or aa, so the total number of sequences that is still possible is $96+64+16=176$ (the last three classes in the above overview). The allowed sequences you want are the sequences that have exactly $2$ AA or aa, because then the other two have to be Aa or aA, so this number is $96$. In the same way as the boy girl problem, the probability now is $\frac{96}{176}$. Note that $\frac{96}{176} = \frac{6}{11}$, which is what Ward found too, but with a slightly different approach. I've added this answer because I thought it might give you a little more insight.

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  • $\begingroup$ Let me repeat that I think that the answer that makes the most sense is $1/4$. You say that "The information we have is that our sequence has at least 2 AA or aa." which is in my opinion a bit of a weird interpretation of "Now two children have been tested and had positive results." $\endgroup$ – Ward Beullens Jan 7 '18 at 16:47
  • $\begingroup$ Well, if they both have tested positive, then under the assumption that the test is 100% accurate, they are both ill. The definition of ill here is that they have genes AA or aa. Then in our sequence of four, we know that at least two elements are AA or aa, one for every ill child. $\endgroup$ – Abby Jan 7 '18 at 17:40
  • $\begingroup$ Thank you for the answer, it is an intuitive road I also followed but as I stated the "answer" of $\frac{36}{64}$ makes me wonder what logic they followed to obtain their answer. $\endgroup$ – Gijsv Jan 7 '18 at 17:41
  • $\begingroup$ Yeah, I have been curious about that as well. I tried to come up with something related to Bayes, but couldn't think of anything yet. If I find something later I'll let you know! $\endgroup$ – Abby Jan 7 '18 at 17:55
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The boy girl paradox shows that these questions are sometimes up for interpretation. Though in this case I believe the most sensible answer is just $1/4$. The question clearly states that only 2 persons are tested, so the other 2 persons still each have a $1/2$ probability of testing either way.

However, if the question stated that all children were tested and that at least two of them tested positive, then the answer would be different.

By enumerating all $16$ possibilities and discarding the $5$ cases in which less than $2$ children test positive we are left with $11$ possibilities. Out of these possibilities there are $\binom{4}{2} = 6$ in which the other children test negative, so the probability would be $6/11$.

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