This seems to be Bayes, but then you cannot have 2 diseases at once.

That should be a partition and I don't see it here.

Let P=psoitive and D1=Disease 1, D2=Disease 2 and D3=Disease 3.

P(D1)=P(D2)=P(D3)=1/3 our partition?

P(P)=P(P)P(P|D1)+P(D2)P(P|D2)+P(D3)P(P|D3)=[.8+.6+.4]/3=.6

Thus P(D1|P)=P(D1P)/P(P)=(.8)(1/3)/.6=4/9.