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I have an assignment question that I just cannot crack.

Show that

$$P(A \cap B) \ge 1-P(\overline{A}) - P(\overline{B})$$

By using the following elementary properties of probabilities:

$P(A) + P(\overline{A}) = 1 \tag{1}$

$P(A) \le 1 \tag{2}$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) \tag{3}$

My progress so far:

Using (1):

$$ P(A \cap B) \ge P(A) + P(B) - 1 $$

Using (2):

$$ P(A) + P(B) \le 2 - 1 $$ $$ P(A) + P(B) \le 1 $$

Using (3):

$$ P(A \cup B) \le 1 $$ $$ P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

Is this sufficient proof that $P(A \cap B) \ge 1-P(\overline{A}) - P(\overline{B})$?

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    $\begingroup$ Hint: Add and subtract 1 from $P(A)+P(B)-1$ and regroup. $\endgroup$ – Dilip Sarwate Mar 11 '18 at 11:27
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Remark about:

$$P(A)+P(B) \le 1.$$

This inequality doesn't makes sense, what if $P(A)=P(B)=1$.

Hint to solve the problem:

Notice that the left has side is just $$1-P(A^c\cup B^c) $$

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