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If P(A) = .75 and P(B) = .6, Prove that $$P(A \cap B) \ge 0.35$$

I have no clue how to do this. I don't want the proof, just so I can copy it, I just need some sort of indication of how to do this.

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    Hint: $P(A \cup B)$ = $P(A) + P(B)$ $-$ $P(A \cap B)$ – 2017-01-20
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    I know, but I don't see how that helps if I don't know P(A U B) – 2017-01-20
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    But you do know that $0\leq Pr(A\cup B)\leq 1$ and that is enough to reach a conclusion. – 2017-01-20

2 Answers 2

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Since $P(A\cup B)\le 1$,

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

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In case the other answer and comments were not already clear enough:

$0\leq Pr(A\cup B)\leq 1$

$(\star)~~~0\leq Pr(A)+Pr(B)-Pr(A\cap B)\leq 1$

$0\leq 0.75 + 0.6 - Pr(A\cap B)\leq 1$

$~$

$0\leq 1.35 - Pr(A\cap B)\leq 1$

$~$

$-1.35\leq -Pr(A\cap B)\leq -0.35$

$~$

$1.35\geq Pr(A\cap B)\geq 0.35$

Note that $(\star)$ is justified by the principle of inclusion-exclusion and the first line is justified by the axioms of probability. Everything else is just algebraic manipulation and arithmetic.