Suppose A and B are two events, not independent. Is the probability P(A and B) > 1/3?
A. P(A)=0.8 and P(B)=0.7
B. P(A or B) = 0.9
A. P(A)=0.8 and P(B)=0.7
B. P(A or B) = 0.9
If A and B are independent, then P(A and B) = P(A) * P(B)
If A and B are NOT independent, then P(A and B) = P(A) * P(B|A). (The second term is "the probability of B given that A happens", and can be written as P(B) * P(A|B) / P(A).)
S1:
We know that the probability of both A and B is LESS THAN OR EQUAL TO the probability of only the less likely one happening. (Getting A and B is less likely than getting B.) So
P(A) > P(B) ≥ P(A and B)
We also know that P(A and B) is MORE LIKELY than P(A) + P(B) - 1. This is because we know that
1 ≥ P(A or B) = P(A) + P(B) - P(A and B)
which we can rearrange by writing
1 ≥ P(A) + P(B) - P(A and B)
and adding P(A and B) to both sides, and subtracting 1 from both sides:
P(A and B) ≥ P(A) + P(B) - 1
So we have the inequality:
.7 ≥ P(A and B) ≥ (.8 + .7 - 1)
.7 ≥ P(A and B) ≥ .5
So P(A and B) ≥ .5, which is > (1/3): SUFFICIENT!
S2:
We don't know the relationship between A and B here. For instance, suppose A and B are not independent in one sense: if A happens, then B only has a 10% chance of happening, and if A doesn't happen, then B has a 0% chance of happening. In this case, P(A and B) < (1/3). But if A and B are "almost independent" (that isn't logically right, but you get the idea), then it's easy to generate numbers in which P(A or B) = 0.9 and P(A and B) > (1/3). So we don't know.
Algebraically, we don't have great limits here. We know from the previous explanation that
P(A) + P(B) - P(A and B) = P(A or B)
so
P(A) + P(B) - P(A and B) = 0.9
but this doesn't allow us to isolate any further.
