# Which one of these options is false?

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.

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.
• 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. – Did Mar 26 '17 at 10:47