Let $A$ and $B$ be events. Is it necessarily true that $\frac{P(A \cap B)}{P(A)} = \frac{P(A \cap B)}{P(B)}$?

Question

Let $A$ and $B$ be events. Is it necessarily true that $\dfrac{P(A \cap B)}{P(A)} = \dfrac{P(A \cap B)}{P(B)}$?

If so, explain why. If not, give a counterexample.

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I'm having a hard time proving this problem. Where should I start? How should I explain it in a way that makes sense and is a valid proof at the same time? Listing random events wouldn't help since it must be true for all events. Can someone walk me through this? Thanks!

Answer 1

Hint:

$P(A|B)$ need not be equal to $P(B|A)$.

Let $A$ be the event that you get number $2$ from a dice toss.

Let $B$ be the event that you get an even number from the same dice toss.

Answer 2

Say you have a standard six-sided die and a fair coin. You roll first the die and get $1,2,3,4,5,6$ with equal prob. and then throw the coin and get $H,T$ with equal prob.

Define $A$ to be "get a $1$ with the die" and $B$ "get a $Head$ with the coin".

Clearly $P(A) = 1/6$ and $P(B) = 1/2$ and both event are independent. Therefore $$ \frac{P(A \cap B)}{P(A)} = \frac{1/12}{1/6} \neq \frac{1/12}{1/2} = \frac{P(A \cap B)}{P(B)} $$

Answer 3

Since $$P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)$$ We can see $$\frac{P(A\cap B)}{P(A)}=\frac{P(A)P(B|A)}{P(A)}=P(B|A)$$ but $$\frac{P(A\cap B)}{P(B)}=\frac{P(B)P(A|B)}{P(B)}=P(A|B)$$ Hence, $$\frac{P(A\cap B)}{P(B)}\neq\frac{P(A\cap B)}{P(A)}$$

The example is when $A$ and $B$ are independent. In such case $P(A|B)=P(A)$ but $P(B|A)=P(B)$

Answer 4

Honestly, i think it does not(i might be wrong, but here we go). Notice that $\frac{P(A \cap B)}{P(A)} $ is actually the probability of the event $B$ knowing that the event $A$ happened. (Here, your sample space is actually the event $A$, so this formula is nothing but $P(C) = \frac{C}{S}$ where $C$ is an event and $S$ the sample space. Now, think about it: Is the probability of an event $A$, knowing that $B$ has happened, equal to the probability of an event $B$, knowing that $A$ happened? The answer is no when $P(A)$ is not equal to $P(B)$. And that can be easily demonstrated: You have: $$ \frac{P(A \cap B)}{P(A)} = \frac{P(A \cap B)}{P(B)} $$ Obviously, this equality only holds if $$P(A) = P(B) $$. An example: I have two dices, one black an one white. Let A be the event of the black resulting in 6 and B the event of the white resulting in 6. I throw the first dice, so i have: $$P(A) = 1/6 $$ And then: $$P(B) = 1/6 $$ Since both of the events are independent, we have that $$P(A \cap B) = P(A)P(B) $$ So: $$P(A/B) = P(A) = P(B/A) = P(B) $$. This is an example that the equality holds for $P(A) = P(B)$. Any case that this doesn't apply, the equality doesn't apply.