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Let $P(A)=0,3$. Only one of the following events has a probability less than $0.3$. Which one?

$A)A \cup B$

$B)\bar{A} \cup B$

$C)A \cap B$

$D)\bar{A} \cup \bar{B}$

Lets check one by one.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ and then $\space 0 \leq P(A \cap B)\leq 0.3 \space$. This happen because, if $\space A \subset B \space$ then $\space P(A \cap B) =P(A) \space$ and so $\space P(B) \geq P(A) \space$. In other hand, if $\space P(A \cap B)=0 \space$ then $\space P(A \cup B)=P(A)+P(B) \space$, and so $\space P(A \cup B) \geq 0.3 \space$ because $\space A \subset B \space$.

If $\space B \subset A \space$, then$\space P(A \cap B)\leq 0.3 \space$, and so $\space P(A \cup B) \geq 0.3 \space$, because $\space P(A)=0.3 \space$ and $\space P(B)\leq P(A) \space$.

Now the second one:

$P(\bar{A} \cup B)=P(\bar{A})+P(B)-P(\bar{A} \cap B)$. It's known that $ \space P(\bar{A})=0.7 \space$, so $\space 0 \leq P(\bar{A} \cap B)\leq 0.7$. If $\bar{A} \subset B$ then $P(\bar{A} \cap B)=P(\bar{A})$ and so $P(\bar{A} \cup B)=P(B)$, wich is $\geq P(\bar{A}) \space$. If $P(\bar{A} \cap B)=0$ then $P(\bar{A} \cup B) \geq0.7$ because $P(\bar{A})=0.7$.

If $B \subset \bar{A}$ then $P(\bar{A} \cap B)=P(B)$, and so $P(\bar{A} \cup B)=P(\bar{A})$.

The third one:

If $A \subset B$ then $P(A \cap B)=P(A)$. But if $B \subset A$ then $P(A \cap B)=P(B)$, and in this case $P(B) \leq P(A)$. So $0 \leq P(A \cap B)\leq 0.3$.The third one would be the right one

This was a question test, without solutions.Can you see if my answer is the right one?Thanks

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$\TeX$ assumes the use of decimal points rather than decimal commas. If you use a comma, it gets interpreted as punctuation between the numbers and spaced accordingly. Being from Germany, where decimal commas are also used, I can sympathize, but in global mathematical notation the decimal point has become the de facto standard. –  joriki Apr 13 '12 at 17:26

2 Answers 2

up vote 1 down vote accepted

Your answer is the correct one, but I'm not sure about your proof. Where does it say that either $B \subseteq A$ or $A \subseteq B$? In any case, $ A \cap B \subseteq A$, so $P(A \cap B) \leq P(A)$. To see this, you can write $A = (A \cap B) \cup (A \setminus B)$, which is a disjoint union, so $P(A) = P(A \cap B) + P(A \setminus B)$. (Hence we have a strict inequality if and only if $P(A \setminus B) \neq 0$.)

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I know that my proof is not quite explicit.As I was thinking, I wrote on the paper. –  João Apr 13 '12 at 17:55
1  
@João: I think that by the word explicit, you mean complete. It is a perfectly fine proof under the assumptions listed (and making such assumptions also makes perfect sense if all you want to do is rule out other alternatives). I was merely pointing out that it isn't a complete proof for the inequality $P(A \cap B) \leq 0.3$ for all events $B$. –  Martin Wanvik Apr 13 '12 at 18:25

That is correct.

Note that B and D can be immediately ruled out because $P(\overline{A}) = .7$ and thus, since $\overline{A} \subseteq (\overline{A} \cup B), (\overline{A} \cup \overline{B})$, we immediately have that $P(\overline{A}) \leq P(\overline{A} \cup B), P(\overline{A} \cup \overline{B})$.

Similarly, because $A \subseteq (A \cup B)$, we have $P(A) \leq P(A \cup B)$, and so option A can also be ruled out, leaving C as the solution.

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