# Are these probabilities independent or not?

P(A) = P(The world would be destroyed today by a meteor) = 1/2

P(B) = P(My pet plant would sprout tomorrow) = 1/2

When combined, there are 3 possible cases.

1. The world will be destroyed, and my plant will die too = 2/4
2. The world will not be destroyed, but my plant still won't sprout = 1/4
3. The world will not be destroyed, and my plant will sprout = 1/4

The point I am confused is the part where you compare P(A∩B) and P($$A^c$$∩B).

So like, P(A∩B) = 0 ≠ P(A)×P(B), which means A and B are not independent.

But, P($$A^c$$∩B) = 1/4 = P($$A^c$$)×P(B), which means A and B are independent.

What am I missing here?

The tricky part is that your probability for $$B$$ is not a straight-up probability, it's a conditional probability. It's conditioned on the fact that $$A$$ does not occur, so really, your probability for $$P(B)$$ is $$P(B|A^C)=0.5$$.

We can see that if you fill out your diagram. $$P(A\cap B)=0$$, so the top left should be $$0$$. $$P(A\cap B^C)=0.5$$, so the bottom left is $$0.5$$. The right hand column should have $$0.25$$ in each box. Therefore, when you add up the probabilities in the top row (to get the probability $$B$$), you get $$0.25$$, which means $$P(A^C\cap B)=0.25\neq 0.5\cdot 0.25$$.

You stated that

P(B) = P(My pet plant would sprout tomorrow) = 1/2

That is not P(B), instead, that is P(B|A$$^c$$) as you have made the assumption that the earth was not destroyed.

If you simulated these set of events many times, you would find P(B) = 1/4.

This would show dependence as:

P(A$$^c$$) x P(B) = 1/2 x 1/4 = 1/8 and hence,

P(A$$^c$$∩B) = 1/4 ≠ P(A) × P(B)

The product rule for dependent events is P(A$$^c$$∩B) = P(A$$^c$$) x P(B|A$$^c$$). This is the formula that you used without realisation as you confused P(B) and P(B|A$$^c$$). That is why you thought that formula showed the probabilities were independent.