4
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Given two independent events $A$ and $B$, with given conditions:
$0 \lt P(A) , P(B) <1 $.
Which one of the following options is/are false?

  1. $A$ and $B’$ are independent.
  2. $A’$ and $B’$ are independent.
  3. $P(A|B) = P(A|B’)$
  4. For any event c, with $0 \lt P(c) \lt 1$, $P(AB|c)= P(A|c)\cdot P(B|c)$

Here is what I tried:

  1. A and B are independent iff: $P(A \cap B)$ $=$ $P(A)\cdot P(B)$
    Now, we have : $P(A) = P(A \cap B) + P(A \cap B')$
    So,
    $ P(A \cap B')$ $=$ $P(A) - P(A \cap B)$
    $=$ $P(A) - P(A)\cdot P(B)$
    $=$ $[1-P(B)]\cdot P(A)$
    $=$ $P(A)\cdot P(B')$
    Thus, 1 is true.

  2. We know that, $P(A’ \cap B’) =P(A \cup B )’$
    $ =1 - P(A \cup B)$
    $ =1 - P(A) - P(B) + P( A \cap B)$
    $ =1 - P(A) - P(B) + P(A)\cdot P(B)$
    $ = [1-P(A)] \cdot [1-P(B)]$
    $ =P(A’)P(B’) $
    Thus, 2 is also true.

  3. By conditional probability, $P(A | B)$ $=$ $\frac{P(A \cap B)} {P(B)}$
    $=$ $\frac{P(A)\cdot P(B)}{P(B)}$
    $=$ $P(A) $
    And
    $P(A | B') $ $ =$ $\frac{P(A \cap B')}{P(B')}$
    $=$ $\frac{P(A)\cdot P(B')}{P(B')}$
    $=$ $P(A) $


The problem is with 4. I tried to disprove it, by finding a counter-example, and I couldn't.


What is the correct answer?

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Is option 4. true?

No : Assume that $C = A\cup B$, then $P(C)\geqslant P(A)>0$ and $$P(AB \mid C)=\frac{P((A \cap B) \cap C)}{P(C)}=\frac{P(A \cap B)}{P(C)} = \frac{P(A)P(B)}{P(C)}$$ while $$P(A\mid C)P(B\mid C) = \frac{P(A \cap C)}{P(C)}\frac{P(B \cap C)}{P(C)} = \frac{P(A)P(B)}{P(C)^2}$$ hence the equality in option 4. holds if and only if $P(C)=1$. Now, $A$ and $B$ are independent hence $$P(C)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$$ Thus, $P(C)=1$ would mean that $$0=1-(P(A)+P(B)-P(A)P(B))=(1-P(A))(1-P(B))$$ which does not hold since $P(A)\ne1$ and $P(B)\ne1$. Thus, for $C=A\cup B$, $0<P(C)<1$ as required for a counterexample to option 4.

To sum up, option 4. never holds.

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  • 1
    $\begingroup$ I think when we have counterexamples to show with independent events, we usually take dice rolls as a starting point. You are welcome! Also, the early down vote seems to have been cancelled out. $\endgroup$ – астон вілла олоф мэллбэрг Mar 26 '17 at 10:29
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    $\begingroup$ This answer gives an example of a probability space $\Omega$ and events $(A,B,C)$ such that $A$ and $B$ are independent, $0<P(A)<1$, $0<P(B)<1$ and $P(A\cap B\mid C)\neq P(A\mid C)P(B\mid C)$. A more ambitious (and more interesting) result is that, for every probability space $\Omega$ and every independent events $A$ and $B$ such that $0<P(A)<1$ and $0<P(B)<1$, there exists some event $C$ such that $P(A\cap B\mid C)\neq P(A\mid C)P(B\mid C)$. Actually, it seems that this is what a full answer to the question as it is formulated, requires. $\endgroup$ – Did Mar 26 '17 at 10:47
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    $\begingroup$ There is nothing to transfer to anything, the question is clearly stated and my last comment explains why, in my opinion, a specific example does not answer it (after all, there might exist some Omega and independent (A,B) such that 4 holds). You can probably get an idea of the abstract construction that allows to prove that 4 never holds if you read carefully the thread of comments on main, where I rather transparently points at the solution... $\endgroup$ – Did Mar 26 '17 at 10:53
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    $\begingroup$ @ABcDexter See the improved answer. Turns out the question wasn't too difficult! $\endgroup$ – астон вілла олоф мэллбэрг Mar 26 '17 at 11:08
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    $\begingroup$ Revamped your post, trying to focus on the relevant points. If this does not suit your taste, please revert to a previous version. (+1) $\endgroup$ – Did Mar 26 '17 at 11:30

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