I think I have a more serious problem regarding these formulas. If a and b goes to 1, regardless of c, Pr(p) and Pr(q) goes to 1. So if p is the statement “q is true.” and q is the statement “p is true.” then p and q must be true, which I think is nonsense. But I cannot see where my mistake is. Could you help please?
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.
I think I have a more serious problem regarding these formulas. If a and b goes to 1, regardless of c, Pr(p) and Pr(q) goes to 1. So if p is the statement “q is true.” and q is the statement “p is true.” then p and q must be true, which I think is nonsense. But I cannot see where my mistake is. Could you help please?
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.