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$A, B$ are independent events.

Then $P(A \cup B)= P(A)+P(B)-P(A)P(B).$

I'm getting stuck on the $P(A) \geq \frac{1}{2}$ mostly because I have never dealt with inequalities.

How should I tackle this type of question?

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    Construct a simple example: an urn contains three balls, one red, one green, one blue, one cyan. The sample space is: $S = \{r, g, b, c\}$. The experiment consists in picking a ball, all three balls being equiprobable. Let $A$ = the event "the chosen ball is either red or green" = $\{r, g\}$, and $B$ = the event "the chosen ball is either blue or green" = $\{b, g\}$. Each of these events has probability $2/3$, which is $\geq 1/2$. Can you calculate $P(A \cup B)$? Can you construct other examples in an attempt to get $P(A \cup B) < 3/4$? – 2017-02-15
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    all three balls are quiprobable? but you have 4 (r,g,b,c) – 2017-02-15
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    @avs what on earth is the point of that – 2017-02-15
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    @SlugPue: the point is to explain the idea rather than carrying out "blind" algebraic manipulations. – 2017-02-15
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    @JackD'Aurizio explaining the idea would look a whole lot differently than start dealing with balls and whatnot – 2017-02-15
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    @amWhy Why you fixed my fixing? Explain please. Thank you! – 2017-02-15
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    @guestuser46, you are right, my typo. (I originally wrote 3 balls.) So, there should be 4 balls, equiprobably chosen. Each of $A, B$ has probability $1/2$. – 2017-02-15
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    @SlugPue, how would you explain the idea? I was trying to lead the OP to *discovering* the idea. All discoveries in mathematics are made experimentally and intuitively. The formal write-ups of formulations and proofs are the aftermath.:) – 2017-02-15
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    @SlugPue http://pauli.uni-muenster.de/~munsteg/arnold.html – 2017-02-15
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    @avs see my answer to he problem. This is a matter of opinion, of course, but there is a reason we do mathematics in the first place: namely to formalize things for the purpose clarity and simplicity. Intricate examples with special cases is the opposite of clarity and simplicity. – 2017-02-16
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    @SlugPue, elegant and instructive, but you are employing certain properties of continuous functions, with which the OP may be unfamiliar. Yes, the ultimate goal is to state clearly, but I was talking about the process of achieving that. Have you looked at the link I posted? Vladimir Arnol'd was a world-class mathematician, a Wolf prize laureate, who also refused (or was pressed to refuse?) to accept the Fields medal. In short, he does have some mathematical credentials. – 2017-02-16
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    @avs very interesting read – 2017-02-16

4 Answers 4

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First, it is clear that for $\mathbb P (A) = \mathbb P(B) = \frac{1}{2}$, you have $\mathbb P(A \cup B) = \frac{3}{4}$. It is easy to see that by increasing either $\mathbb P (A)$ or $\mathbb P (B)$ you increase $\mathbb P(A \cup B)$, because for $x, y \in [0, 1]$: $$ f(x, y) = x + y - xy \Rightarrow\\ \frac{\partial f}{\partial x}(x, y) = 1 -y \geq 0 $$

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hint:

$(x+y-xy-1) = (x-1)(1-y)$

And if $1 \geq x,y \geq \frac{1}{2}$ then $|x-1||y-1| \leq \frac{1}{4}$

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$P(A\cup B)=1-(1-P(A))(1-P(b))\geq\frac{3}{4}$

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    Nice, but how instructive is this? Does it show the OP how they could have approached the problem to discover the solution themselves? – 2017-02-15
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    What happened to the $P(A)+P(B)-P(A)P(B)$ – 2017-02-15
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    @ guestuser46 $1-(1-P(A))(1-P(B))\geq1-\left(1-\frac{1}{2}\right)(1-P(B))\geq1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2}\right)=\frac{3}{4}$ – 2017-02-15
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    @avs See my previous comment. – 2017-02-15
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    You have 1 because $(P(A)+P(B)=1)$ right? – 2017-02-15
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    @MichaelRozenberg, it's all correct, but what intuitive considerations would have led the OP to *discover* how to solve the problem? I have a feeling they weren't even in the habit of thinking of events as subsets of the sample space. Look at their last comment: for some reason, they think $A$ and $B$ are complementary. – 2017-02-15
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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,

$$\frac{3}{4}\le P(A)+P(B)-P(A)P(B)\le 1$$

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    How did you get $$P(A)P(B)\le \frac{(P(A)+P(B))^2}{4}\to -P(A)P(B)\ge -\frac{(P(A)+P(B))^2}{4}$$ – 2017-02-15
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    that is AM-GM inequality: $$\frac{a+b}{2}\ge \sqrt{ab} \Leftrightarrow ab\le \frac{(a+b)^2}{4}$$ – 2017-02-15
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    Yeah I got the formula, so you got a positive and negative since its $\sqrt{ab}$. – 2017-02-15
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    @guestuser46: take the last inequality and multiply by $-1$ – 2017-02-15
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    May I ask why multiply by $-1$? – 2017-02-15
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    Because in your formula you have $-P(A)P(B)$ – 2017-02-15
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    I did not see that, now I'm starting to get it thanks! – 2017-02-15
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    one last question how did you get $x-\frac{x^2}{4}$? – 2017-02-15
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    I just used $x= P(A)+P(B)$. – 2017-02-15