Hallo zusammen

Stimmt mein Ansatz für Aufgabe a und b?

The colour of a person’s eyes is determined by a unique pair of genes. If both are the genes

for blue eyes, the person will have blue eyes; if they are both the genes for brown eyes, the person will

have brown eyes; if one is a blue-eyed gene and the other brown-eyed, the person will have brown eyes

(i.e., the brown-eyed gene is dominant). A newborn baby receives one eye gene independently from each

parent, and the gene he receives from each parent has an equal probability of being either one of that

parent’s two eye genes. Suppose that Xavier and both his parents have brown eyes, but Xavier’s sister

has blue eyes.

Gegeben: 

BB -> B

BrBr -> Br

BBr -> B

a) What is the probability that Xavier has one blue-eyed gene?

Entweder kann Xavier folgende Genen für ein blaues Auge haben 0 Gen, 1 Gen oder 2 Gen.

Also folgende Kombis gibt es

P(Br,Br)

P(B,Br)

P(B,B)

2/3 ist die WS, da nur von 3 Fällen zwei möglich sind oder?

b) If Xavier’s wife has blue eyes, what is the probability that their first child will have blue eyes?

Also 

P(BB) = Xavier hat P(Br,B) oder P(B,Br) wie berechnet man hier die WS?

Vielen Dank für eure Hilfe!

Liebe Grüsse

Sayuri