# Conditional probability prove or disprove

• MHB
• mathmari
In summary, the conversation discusses two statements about conditional probability and their validity. The first statement, $P(A\mid B)=1-P(\overline{A}\mid B)$, is proven to be true by using the definition of conditional probability and the complement rule. The second statement, $P(A\mid B)=1-P(A\mid \overline{B})$, is shown to be false by providing a counterexample and using proof by contradiction. The conversation also delves into the definition of conditional probability and how it is used in the proofs.
mathmari
Gold Member
MHB
Hey!

Let $P$ be a probability measure on a $\sigma$-Algebra $\mathcal{A}$. I want to prove or disprove the following statements:
1. $P(A\mid B)=1-P(\overline{A}\mid B)$, for $A, B\in \mathcal{A}$
2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$
I have done the following:
1. We have that $B=\left (A\cap B\right )\cup \left (\overline{A}\cap B\right )$. Since the sets $\left (A\cap B\right )$ and $\left (\overline{A}\cap B\right )$ are disjoint we have that $P\left [\left (A\cap B\right )\cup \left (\overline{A}\cap B\right )\right ]=P\left (A\cap B\right )+P \left (\overline{A}\cap B\right )$.
Therefore, we get $P(B)=P\left (A\cap B\right )+P \left (\overline{A}\cap B\right )$.

We have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}\Rightarrow P(A\cap B)=P(A\mid B)P(B)$ and $P(\overline{A}\mid B)=\frac{P(\overline{A}\cap B)}{P(B)}\Rightarrow P(\overline{A}\cap B)=P(\overline{A}\mid B)P(B)$.

So, we get $P(B)=P(A\mid B)P(B)+P(\overline{A}\mid B)P(B) \Rightarrow P(B)=P(B)\left [P(A\mid B)+P(\overline{A}\mid B)\right ] \overset{P(B)>0}{\Rightarrow }1=P(A\mid B)+P(\overline{A}\mid B) \Rightarrow P(A\mid B)=1-P(\overline{A}\mid B)$
So, the statement is true.

Is this correct? 
2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$

From the definition of conditional probability then we get the statement $\frac{P(A\cap B)}{P(B)}=1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$.

For $A=\Omega$ we get $\frac{P(\Omega\cap B)}{P(B)}=1-\frac{P(\Omega\cap \overline{B})}{P(\overline{B})}\Rightarrow \frac{P( B)}{P(B)}=1-\frac{P( \overline{B})}{P(\overline{B})}\Rightarrow 1=1-1 \Rightarrow 1=0$, a contradiction.

Therefore, this statement is in general not true.

Is this correct?

Last edited by a moderator:
mathmari said:
1. $P(A\mid B)=1-P(\overline{A}\mid B)$, for $A, B\in \mathcal{A}$

Hey mathmari!

What if $P(B)=0$? (Wondering)

mathmari said:
2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$

From the definition of conditional probability then we get the statement $\frac{P(A\cap B)}{P(B)}=1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$.

How do we get that? (Wondering)

I like Serena said:
What if $P(B)=0$? (Wondering)

If $P(B)=0$ then the conditional probaibility $P(A\mid B)$ is not defined, is it? (Wondering)
I like Serena said:
How do we get that? (Wondering)

By definition we have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\mid \overline{B})=\frac{P(A\cap \overline{B})}{P(\overline{B})}$, or not? Therefore we get that $P(A\mid B)=1-P(A\mid \overline{B})$ is equivalent to $1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$. Is this wrong? (Wondering)

mathmari said:
If $P(B)=0$ then the conditional probaibility $P(A\mid B)$ is not defined, is it? (Wondering)

Ah yes.
That's because $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, which is indeed undefined if $P(B)=0$.

mathmari said:
By definition we have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\mid \overline{B})=\frac{P(A\cap \overline{B})}{P(\overline{B})}$, or not?

Yes.

mathmari said:
Therefore we get that $P(A\mid B)=1-P(A\mid \overline{B})$ is equivalent to $1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$. Is this wrong?

Well, I'm only aware of the complement rule $P(\overline U)=1-P(U)$, but it's not clear to me how this is applied.

I like Serena said:
Well, I'm only aware of the complement rule $P(\overline U)=1-P(U)$, but it's not clear to me how this is applied.

From the definition of the conditional probability we have that $P(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$ and $P(A\mid \overline{B})=\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Replacing these at $\mathbb{P}(A\mid B)=1-\mathbb{P}(A\mid \overline{B})$ we get $\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=1-\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Isn't it correct? (Wondering)

mathmari said:
From the definition of the conditional probability we have that $P(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$ and $P(A\mid \overline{B})=\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Replacing these at $\mathbb{P}(A\mid B)=1-\mathbb{P}(A\mid \overline{B})$ we get $\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=1-\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Isn't it correct? (Wondering)

Oh wait! You're doing a proof by contradiction!
I totally missed that.
Perhaps good to mention that. That is, 'suppose it is true, then...'. (Nerd)

I like Serena said:
Oh wait! You're doing a proof by contradiction!
I totally missed that.
Perhaps good to mention that. That is, 'suppose it is true, then...'. (Nerd)

Ah ok! Is the counterexample that I used correct? (Wondering)

mathmari said:
Ah ok! Is the counterexample that I used correct?

Yes. (Nod)

I like Serena said:
Yes. (Nod)

Great! Thank you! (Yes)

## 1. What is conditional probability?

Conditional probability is a mathematical concept that measures the likelihood of an event occurring given that another event has already occurred. It is represented by P(A|B), where A and B are events.

## 2. How is conditional probability calculated?

The formula for calculating conditional probability is P(A|B) = P(A and B) / P(B). This means that the probability of event A occurring given that event B has occurred is equal to the probability of both events A and B occurring divided by the probability of event B occurring.

## 3. Can conditional probability be used to prove or disprove causation?

No, conditional probability only measures the likelihood of an event occurring given that another event has already occurred. It does not prove or disprove causation between the two events.

## 4. What is the difference between conditional probability and joint probability?

Conditional probability measures the likelihood of one event occurring given that another event has occurred, while joint probability measures the likelihood of two events occurring together. Conditional probability uses the formula P(A|B) while joint probability uses the formula P(A and B).

## 5. How is conditional probability used in real life?

Conditional probability is used in many fields, such as statistics, economics, and medicine. It can be used to determine the likelihood of a patient having a certain disease given their symptoms, or the probability of a stock price increasing given market trends. It is also used in everyday decision making, such as the probability of getting into a car accident given the weather conditions.

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