# Probability - How to determine if independent of dependent

I have 2 events, $A$ and $B$ which are in a space $S$, such that $P(A')=0.3$, $P(B)=0.5$ and $P(A' \bigcup B')=0.7$.

I need to answer a series of questions based upon this information, but I'm running into a brick wall almost immediately.

Let's say I need to find $P(A \bigcup B)$. As far as I can tell, the formula for this depends on whether it's independent or dependent.

However it's not until the final question that I'm asked whether A and B are independent. Do I not need to know this before I answer any of the previous questions? Where do I even begin here?

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Well, you can compute the union with $$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$ That is, the probability of the whole is the sum of the probabilities of the parts, but dropping the "overlaps" so they are not counted twice.

Note that you can get $P(A \cap B)$ from $P(A^c \cup B^c)$, which is the data you are provided with.

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Can you explain how to get P(A∩B) from that? I have no idea how to proceed because every resource I look at has two methods depending on whether they're dependent or independent, which I don't know. – karoma Oct 13 '12 at 18:41
You have to use de Morgan's Laws, which state that $A^c \cup B^c = (A \cap B)^c$ and the other way around. – busman Oct 13 '12 at 18:42
karoma: What are the resources you are looking at? – Did Oct 13 '12 at 18:43
Bingo. 3 hours of searching and I never saw that mentioned. Thank you kindly. – karoma Oct 13 '12 at 18:44
Draw a picture (Venn Diagram). The world will be split up into $4$ parts. Using the information supplied, you can figure out the weights of the individual parts, and then you can answer any question. – André Nicolas Oct 13 '12 at 18:59

This kind of problem is often solved quite easily by drawing a picture:

Here $A'$ is the green together with the white, $B$ is the green together with the blue, and $B\,'$ is the red together with the white, so $A'\cup B\,'$ is the green, white and red $-$ i.e., everything except the blue. Since $P(A'\cup B\,')=0.7$, you can immediately conclude that the blue is the other $0.3$, and since the blue is $A\cap B$, we have $P(A\cap B)=0.3$. $P(B)=0.5$, so the green must have probability $0.5-0.3=0.2$; the green is $B\cap A'$, so $P(A'\cap B)=0.2$.

Now $P(A')=0.3$, and $A'$ is the green together with the white; the green is $0.2$ so the white is $0.3-0.2=0.1$. Thus, $P((A\cup B)')=0.1$. The blue, green, and white now account for $0.3+0.2+0.1=0.6$, so the red must be the remaining $1.0-0.6=0.4$: $P(A\cap B\,')=0.4$. In tabular form:

$$\begin{array}{rl} \text{Red:}&P(A\cap B\,')=0.4\\ \text{Blue:}&P(A\cap B)=0.3\\ \text{Green:}&P(A'\cap B)=0.2\\ \text{White:}&P((A\cup B)')=0.1 \end{array}$$

Every possible combination of $A,B,A'$, and $B\,'$ is made up of these four pieces, so from them you can get anything.

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