# Show that P($A^c\cap B^c)\leq (1-\epsilon)^2$ for dependent distinct events A and B such that P(A)$\geq\epsilon$ and P(B)$\geq\epsilon$ [closed]

If two distinct events A and B are dependent such that P(A)$$\geq\epsilon$$ and P(B)$$\geq\epsilon$$, we want to show that P($$A^c\cap B^c)\leq (1-\epsilon)^2.$$

• Consider one flip of a fair coin and let both $A$ and $B$ be the event that a heads appears. Thus, we may set $\epsilon=1/2.$ The event $A^c\cap B^c$ is the appearance of a tails, with probability $1/2.$ You are asking us, therefore, to show that $$1/2 = \Pr\left(A^c\cap B^c\right) \le (1-1/2)^2 = 1/4,$$ which obviously is not true. – whuber Apr 23 '19 at 17:55
• In your example, A and B are independent. But in the question, two events are dependent. – mathfear Apr 23 '19 at 17:59
• In my example, $A$ and $B$ are anything but independent! They are identical and have probability strictly between $0$ and $1.$ Thus, $$\Pr(A)\Pr(B) = 1/4 \ne 1/2=\Pr(A\cap B)$$ shows lack of independence by definition. – whuber Apr 23 '19 at 17:59
• @mathfear I think you misunderstood. In whuber's example, $A$ is the event that a coin flip shows heads, and $B$ is the event that the same coin flip shows heads. Whuber is writing about a single flip, not writing about the outcomes of 2 distinct flips. – Sycorax says Reinstate Monica Apr 23 '19 at 18:19
• @whuber these two events are distinct. I edited the question now. – mathfear Apr 23 '19 at 18:31

We have \begin{align} P(A^c \cap B^c) & = P((A \cup B)^c) \\ & = 1 - P(A \cup B) \\ & = 1 - P(A) - P(B) + P(A \cap B) \\ & \leq 1 - 2 \epsilon + P(A \cap B) \\ & = (1 - \epsilon)^2 - \epsilon^2 + P(A \cap B), \end{align} so a sufficient condition is that $$P(A \cap B) \leq \epsilon^2$$, i.e. the probability of the intersection must be small. An extreme case is when $$A$$ and $$B$$ are disjoint so that $$P(A \cap B) = 0$$. As whuber's comment shows, the result is not true in general and can fail when there is too much overlap between $$A$$ and $$B$$.