You are right. As i understood the problem, the man has one dominant diseased allele "D", and another wilt-type allele "d". Each of these two alleles can be inherited. Whenever "D" it is inherited, the child will also have the disease. When the "d" is inherited, the phenotype of the child depends on the allel it gets from the mother (if mother is "dd" (homozygous wild-type), then the child will be healthy, otherwise is can still get a "D" for the mother and can thus have the disease).

As long as the mother is not DD, these parents can get healthy as well as diseased children. We can not say whether or not the next child will be diseased. For the knowledge we have we can only give probabilities for these events.

Probabilities of two unrelated events happening together (e.g. having two diseased children) is the product of the individual probabilities. This follows from the basic laws of probability calculus. At the other extreme, for perfectly correlated/dependent events, the probability of any set of such events is the same as the individual probability of this event. The twins are such dependent events. All twins have the same genotype. One twin is like a clone of the other. If you now consider one child, you can make an assumption about its genotype (based on probability), abd whatever you find for this individual will be the very same for each clone of this child.

Given the mother is "DD", then all children must be diseased. There is no other option. In this case one can say, with certainty, that all children will be diseased.

Given the mother is "Dd", then children can have the genotypes DD, Dd and dd (1:2:1). If DD is not lethal, there are three times as many possible events leading to a deseased child, so P(child is diseased) = P(twin-pair is diseased) = 3/4. This is than also the chance that a twins-pair is diseased. P(both pairs are diseased) id 3/4*3/4 = 9/16. Note that is is impossible that 1 or 3 children are diseased!

Given the mother is "dd", the children can be Dd or dd (1:1). P(twin-pair is diseased) = 1/2, P(both twin-pairs are diseased) = 1/2*1/2=1/4.

From the given answers A and C are *possible* but not sure, whereas B is *impossible*.

"As long as the mother is not DD, these parents can get healthy as well as diseased children."

The mother was healthy.

"We can not say whether or not the next child will be diseased. For the knowledge we have we can only give probabilities for these events."

"...A and C are *possible* but not sure"

This is the exact answer to my fuzzy feeling. ;)

Thanks!

I don’t know if your question is more about single causation or more about probability.

Is - statements: the problem with existential statements is that they speak of “at least one” thing. We need a special procedure to reduce this to one single thing: "There is at least one thing x with property F and for all things Fy: y = x.” In ordinary language: “There is one and only one thing F”. Example: “There is at least one thing god, we call it x and all other gods y are identical with x.”

Single causation: problem: could be called “irreproducible”. Example: evolution as a whole, the history of the universe etc. Gerhard Vollmer has some arguments for taking single causation into account.

Here I got some information about single causation:

[ http://www.philosophie-wissenschaft-kontroversen.de/suche.php?suche3=Einzelfallverursachung&x=4&y=6 ]

About laws of probability in the context of causation:

[ http://www.philosophie-wissenschaft-kontroversen.de/suche.php?suche3=WschkG&x=3&y=8 ]

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[ http://philosophy-science-humanities-controversies.com/search.php?suche3=WschkG ]

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