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If event $A$ and $B$ are events such that $P(A)$ and $P(B)$ are either $0$ or $1$ and $A$ is subset of $B$, then $A$ and $B$ are dependent events.

Proof: Since $A\subset B$, we have $A\cap B=A$ and so $P(A\cap B)=P(A)$.

$\therefore$ $P(A\cap B)-P(A)P(B)=P(A)-P(A)P(B)=P(A)[1-P(B)]$
Since $P(A)>0$ and $P(B)<1$,

(original image)

Where did that last step come from, involving $P(A \cap B) - P(A)P(B)$? How did the solution come up with that?

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There is only one step. The point is showing that $P(A \cap B) \neq P(A) P(B)$, i.e. that $P(A \cap B) - P(A)P(B) \neq 0$. The steps to $P(A \cap B) - P(A)P(B) = P(A) (1 - P(B))$ are well explained in your writing, but if $P(A) = 0$ or $P(B) = 1$, your events are indeed independent (you will get $0$). So perhaps the theorem you need to assume you don't have such cases happening. If that is of any help... ( I must say it doesn't look quite clear what you assume and what you don't on the values of $P(A)$ and $P(B)$.) –  Patrick Da Silva Jul 23 '12 at 5:14
    
So it's proving by contradiction essentially? –  sidht Jul 23 '12 at 5:16
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@jak: No, the statement you are trying to prove is just wrong. If $P(A) = 0$ or $P(A) = 1$, then $A$ is independent of every event, including itself. Perhaps the "either ... or" was supposed to be "neither ... nor". –  Robert Israel Jul 23 '12 at 5:25
    
One common strategy for showing that $x\ne y$ is to show that $x-y\ne 0$. In this case, it is not really of great help, since from $P(A)=P(A)P(B)$, we can, assuming that $P(A)\ne 0$, cancel and deduce that $P(B)=1$. –  André Nicolas Jul 23 '12 at 5:36
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3 Answers 3

up vote 5 down vote accepted

The result has clearly been misstated or miscopied. The proof is correct for the following result:

If event $A$ and $B$ are events such that $P(A)$ and $P(B)$ are neither $0$ nor $1$, and $A$ is subset of $B$, then $A$ and $B$ are dependent events.

The reason for looking at $P(A\cap B)-P(A)P(B)$ is that by definition, $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$, i.e., if and only if $P(A\cap B)-P(A)P(B)=0$. But the hypothesis that $A\subseteq B$ implies that $P(A)=P(A\cap B)$, so $$P(A\cap B)-P(A)P(B)=P(A)-P(A)P(B)=P(A)\big(1-P(B)\big)\;,$$

which is $0$ if and only if either $P(A)=0$ or $1-P(B)=0$, i.e., if and only if either $P(A)=0$ or $P(B)=1$. These possibilities are ruled out by the corrected version of the hypothesis, so it must be the case that $P(A\cap B)-P(A)P(B)\ne 0$, $P(A\cap B)\ne P(A)P(B)$, and hence by definition $A$ and $B$ are not independent (which of course means that they are dependent).

The answer by copper.hat shows why the stated version is wrong.

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There are only three possibilities to consider:

$$\begin{matrix} PA & PB & P (A \cap B) & PA.PB \\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{matrix}$$

In all cases $PA.PB = P(A \cap B)$, so the events are independent.

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Seems like there's something wrong with the initial statement to me. If $ A = \emptyset \subset B $ then $P(A \cap B)=P(\emptyset \cap B)=P(\emptyset)=0=P(\emptyset)P(B)=P(A)P(B)$.

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