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Suppose $A,B \in F$ are independent eents. If $P(A)\geq \frac{1}{2}$ and $P(B) \geq \frac{1}{2}$ , show that $P( A \cup B) \geq \frac{3}{4}$.

Someone helped me to do this and this is what they got: By AM-GM

$$P(A)P(B)\le \frac{(P(A)+P(B))^2}{4}\to -P(A)P(B)\ge -\frac{(P(A)+P(B))^2}{4}\\ P(A)+P(B)-P(A)P(B)\ge P(A)+P(B)-\frac{(P(A)+P(B))^2}{4}=x-\frac{x^2}{4}$$

where $1\le x=P(A)+P(B)\le 2$

So,

$$\frac{3}{4}\le P(A)+P(B)-P(A)P(B)\le 1$$ My problem is that I got it partially wrong, my question is where is it wrong so I can learn how to do these types of questions.

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    Why not considering the probability of having neither $A$ and $B$ and then compute $1-P(\bar{A} \cap \bar{B})$ ? – 2017-02-27
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    Seems okay, but a little complicated. You were probably dinged for not realizing what should have been known by you that $P(A\cup B) = 1 - \overline {P(A \cup B)} = 1 - P(\overline A \cap \overline B) = 1 - P(\overline A)\times P(\overline B)$ and $P(\overline A) = 1 - P(A) \le 1/2$ so $1 - P(\overline A)\times P(\overline B) \ge 1 - 1/2*1/2 = 3/4$. Far simpler and obvious and it should have been known by you. But ... your way is perfectly fine. Just harder than hell. – 2017-02-27
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    Oh.. okay, thanks for the feedback. – 2017-02-27

1 Answers 1

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Hint. Use $P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - P(\overline A \cap \overline B) = 1 - P(\overline A)P(\overline B)$.