If a=b=1 then Pr(p & q) / Pr(p) = P(p & q) / Pr(q) = 1 so that Pr(p) = Pr(q) = Pr(p & q). That doesn’t require that Pr(p), Pr(q), or Pr(p & q) goes to 1. It just means that in a Venn diagram, p and q coincide (or in popular parlance, “are a circle”).
I used the expression I derived in the post, Pr(q)=bcac−ab+b. However I didn’t notice that c goes to 0 too, at least for the example I gave in my previous comment. So there seems to be no issue as long as c goes to 0 since it causes the indeterminate form 0⁄0.
If a=b=1 then Pr(p & q) / Pr(p) = P(p & q) / Pr(q) = 1 so that Pr(p) = Pr(q) = Pr(p & q). That doesn’t require that Pr(p), Pr(q), or Pr(p & q) goes to 1. It just means that in a Venn diagram, p and q coincide (or in popular parlance, “are a circle”).
How did you get Pr(p) = Pr(q) = 1?
I used the expression I derived in the post, Pr(q)=bcac−ab+b. However I didn’t notice that c goes to 0 too, at least for the example I gave in my previous comment. So there seems to be no issue as long as c goes to 0 since it causes the indeterminate form 0⁄0.