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The Question:

Given three events A,B,C. (For all this question the conditional property is well defined).
a) Knowing that:
$P(A\cap B\cap C)=P(A\cap B)P(C)$
And the above equality being right if we swap any event with it's complement, for example:
$P(A\cap B\cap C^c)=P(A\cap B)P(C^c)$
$P(A\cap B^c \cap C) = P(A\cap B^c)P(C)$
$P(A^c\cap B^c \cap C) = P(A^c \cap B^c)P(C)$ ...

Prove or disprove:
1- Events $A,C$ are independent.
2- Given $A$, the events $B,C$ are independent. (in other words: $P(B\cap C|A)=P(B|A)P(C|A)$).

b) Knowing that $A,C$ are independent, and $B,C$ are independent..
Prove or disprove:
1- Given $A$, the events $B,C$ are independent.
2- $P(A\cap B\cap C)=P(A\cap B)P(C)$


My Work:
a) 1- So the goal here is to check if $P(A\cap C)=P(A)P(C)$.
I'm trying to reach that by writing: $P(A\cap B\cap C) = P(B)P(A|B)P(C|A\cap B)$, I've tried to re-adjust it to:
$P(A\cap C\cap B) = P(A\cap C)P(B|A\cap C)$. now using the complements: $P(A\cap C\cap B^c)=P(A\cap C)P(B^c|A\cap C)$.
And here I got stuck, I didn't know how to reach my goal, tried to find a counter example and it was very long (since I need to care about all the complements) and it didn't work.

2-Again tried to re-adjust again (like in (1)), but I really didn't reach anything interesting that's close to the goal.

b) 1- I know that $P(A\cap C)=P(A)P(C)$ and $P(B\cap C)=P(B)P(C)$. gotta proof that: $P(B\cap C|A)=P(B|A)P(C|A)$.
$P(B|A)P(C|A)=\frac {P(B\cap A)P(C\cap A)}{P(A)^2}=\frac {P(B\cap A)P(C)P(A)}{P(A)^2}= \frac{P(B\cap A)P(C)}{P(A)}$, And here I'm stuck again...

2- Tried to readjust a little and I've no idea how to prove or disprove.


I want to know how off is my work from the solution, I will really appreciate any hints or feedback on my work.
I have tried to think about it intuitively but it was much harder, and I'm wondering if that could help.
Thanks in advance to everyone.

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1 Answer 1

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EDIT: Changed answer based on information that $P(A^c\cap B^c \cap C) = P(A^c \cap B^c)P(C)$ and $P(A\cap B^c \cap C) = P(A\cap B^c)P(C)$


a1.

To prove: $P(A\cap C) = P(A)P(C)$

We have that: \begin{align*} P(A\cap B\cap C)&=P(A\cap B)P(C)\\ P(A\cap B^C\cap C)&=P(A\cap B^C)P(C)\\ P(A\cap B\cap C) + P(A\cap B^C\cap C)&=P(A\cap B)P(C)+P(A\cap B^C)P(C)\tag{1} \end{align*} Since the sets $A\cap B\cap C$ and $A\cap B^C\cap C$ are disjoint, \begin{align*} P(A\cap B\cap C)+P(A\cap B^C\cap C) &= P((A\cap B\cap C)\cup(A\cap B^C\cap C))\\ &= P((A\cap (B\cup B^C)\cap C)\\ &=P((A\cap C)\tag{2} \end{align*} Then from (1) and (2) \begin{align*} P((A\cap C)&=P(A\cap B)P(C)+P(A\cap B^C)P(C)\\ &=(P(A\cap B)+P(A\cap B^C))P(C)\tag{3} \end{align*} But since $A\cap B$ and $A\cap B^C$ are disjoint sets, $P(A\cap B)+P(A\cap B^C) = P((A\cap B)\cup (A\cap B^C))=P(A\cap(B\cup B^C))=P(A)-(4)$

From (3) and (4) $$P((A\cap C)=P(A)P(C)$$


a2. To prove: $P(B\cap C|A)=P(B|A)P(C|A)$ \begin{align*} P(B\cap C|A) &=\frac{P(A\cap B\cap C)}{P{A}}\\ &=\frac{P(A\cap B)P(C)}{P{A}}\\ &=\frac{P(B|A)P(A)P(C)}{P{A}}\\ &=\frac{P(B|A)P(A\cap C)}{P{A}}\\ &=\frac{P(B|A)P(C|A)P(A)}{P{A}}\\ &=P(B|A)P(C|A) \end{align*}


b1. $P(A\cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$

To prove: $P(B\cap C|A)=P(B|A)P(C|A)$

We construct the appropriate venn diagram as a counterexample. Let $x\in (0,\frac{1}{4})\setminus\{\frac{1}{8}\}$

\begin{array}{|c|c|c|c|} \hline A & B& C &\text{Probability of set}\\ \hline 0 & 0 & 0 & \frac{1}{8} \\ \hline 0 & 0 & 1 & x \\ \hline 0 & 1 & 0 & \frac{1}{8} \\ \hline 0 & 1 & 1 & \frac{1}{4}-x \\ \hline 1 & 0 & 0 & \frac{1}{8} \\ \hline 1 & 0 & 1 & \frac{1}{4}-x \\ \hline 1 & 1 & 0 & \frac{1}{8} \\ \hline 1 & 1 & 1 & x \\ \hline \end{array}

Table Explanation: In the example above, the 0's and 1's are to indicate membership in the set. For example $A=0,B=0,C=1$ indicates the set $A^C\cap B^C\cap C$ and $A=1,B=1,C=1$ indicates $A\cap B\cap C$. The 8 rows are then the 8 parts of the venn diagram with three variables. Each row of the table indicates a disjoint subset of the sample space. With this table, one can calculate $\mathbb P [A]$ as $A = (A\cap B\cap C) \cup(A\cap B^C\cap C)\cup(A\cap B\cap C^C)\cup(A\cap B^C\cap C^C)$, then $\mathbb P[A]$ would be the sum of the probabilities of those four rows of the table

Here $\mathbb P(A)=\mathbb P(B)=\mathbb P(C)=\frac{1}{2}$ and $\mathbb P(A\cap C)=\frac{1}{4} = P(A)P(C)$ and $\mathbb P(B\cap C)=\frac{1}{4} = P(B)P(C)$

But $P(B\cap C|A=1) = x \neq P(C|A=1)P(B|A=1) = \frac{1}{2}(\frac{1}{4}-2x)$


b2. Prove: P(A\cap B\cap C)=P(A\cap B)P(C) Same example as above: $P(A\cap B\cap C) = x\neq P(A\cap B)P(C) = (\frac{1}{8} + x)\frac{1}{2}$

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  • $\begingroup$ Thanks for the answer Rahul!, I couldn't quite get the chart with the $P(S)$, are you constructing a counter example with $\Omega={0,1}$? if you could provide extra explanation that would help alot! I appreciate the valuable lesson too! $\endgroup$
    – Pwaol
    Apr 15, 2021 at 18:03
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    $\begingroup$ The 0's and 1's are to indicate membership in the set. For example $A=0,B=0,C=1$ indicates the set $A^C\cap B^C\cap C$. The 8 rows are then the 8 parts of the venn diagram with three variables. Notice that each row is disjoint. Let me add this explanation to the answer also $\endgroup$ Apr 15, 2021 at 18:06
  • $\begingroup$ Thanks! now that part is clear, I'm still struggling to understand the probabilities you wrote under $P(S)$, I assume $S$ is the set of $A,B,C$ in the specific row, but what are these probabilities? what is omega and what events are $A,B,C$? $\endgroup$
    – Pwaol
    Apr 15, 2021 at 18:14
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    $\begingroup$ You can assume any sample space $\Omega$ you like. The only condition is that you define 3 events A B and C over this space (just like how you would construct a venn diagram). The table then gives you a probability measure over the space, which is split into $8=2^3$ disjoint subsets. $\endgroup$ Apr 15, 2021 at 18:19
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    $\begingroup$ Thanks alot! now I understand, that's a really good way of constructing a counter example, I got one last question if possible, when you saw the question, did you see through it that a1 and a2 are false? or was it after you did the diagram and saw it? just trying to learn to how to approach these questions and I want to see what were your initial thoughts that led you to search for a counter example instead of proving. Really appreciate all the help from today and yesterday :). $\endgroup$
    – Pwaol
    Apr 15, 2021 at 18:27

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