Conditional Probability

What is the proof for the following statement?

\begin{gather}
P(A \! \left. \right| B)=\dfrac{P(A \cap B)}{P(B)}
\end{gather}

Posted from TSR Mobile

(edited 10 years ago)

Reply 1

interstitial

OP

18

Bump!

No one at all? :frown:

Posted from TSR Mobile

Reply 2

hassassin04

17

There is no proof. It is the definition of it.

Reply 3

interstitial

OP

18

Original post by hassassin04

There is no proof. It is the definition of it.

It seems very strange to define something like that without laying out any reasoning or steps.

Can you explain how you know that this is the right definition for this and actually means the probability of A given B without having to prove anything?

Reply 4

wofldog

Original post by majmuh24

Bump!

No one at all? :frown:

I no longer study statistics, but the best 'proof' of this formula that I came across was to simply draw a Venn diagram and think about what it is the conditional probability represents. I don't think there exists a proof so much as that formula is the fundamental definition of conditional probability. Nonetheless, I'll try:

Take your two events, A and B, which exist in the same sample space.

Venn 1.png

Now, you want to find

P(A|B)

, or, the probability that A will happen given that B has already occurred. Now there only exists a small region where both A and B can occur simultaneously,

A \cap B

(Sorry, that should say

A \cap B

, not

A \ and \ B

)
However, there is still a probability of B simply occurring again, that means you're calculating the probability of

P(A \cap B)

while also restricting your sample space to

P(B)

.

In simplistic terms, you're trying to calculate the proportion of the area which

A \cap B

takes up inside

B

in your Venn diagram. Replace areas with probabilities and you have your equation!

Hopefully this makes sense (and is correct). If any statisticians want to call me out on anything, feel free!

(edited 10 years ago)

Reply 5

interstitial

OP

18

Original post by wofldog

I no longer study statistics, but the best 'proof' of this formula that I came across was to simply draw a Venn diagram and think about what it is the conditional probability represents. I don't think there exists a proof so much as that formula is the fundamental definition of conditional probability. Nonetheless, I'll try:

Take your two events, A and B, which exist in the same sample space.

Venn 1.png

Now, you want to find

P(A|B)

, or, the probability that A will happen given that B has already occurred. Now there only exists a small region where both A and B can occur simultaneously,

A \cap B

(Sorry, that should say

A \cap B

, not

A \ and \ B

)
However, there is still a probability of B simply occurring again, that means you're calculating the probability of

P(A \cap B)

while also restricting your sample space to

P(B)

.

In simplistic terms, you're trying to calculate the proportion of the area which

A \cap B

takes up inside

B

in your Venn diagram. Replace areas with probabilities and you have your equation!

Hopefully this makes sense (and is correct). If any statisticians want to call me out on anything, feel free!

Thanks for this, your proof was intuitive and easy to follow! Nice work! :biggrin:

Reply 6

hassassin04

17

Example: We toss a coin 3 times. Lets define our sample space( all possible outcomes) as

\Omega=[H,T]^3

. Let event A be that we win the game if we get two heads. Note,

A=[THH,HTH,HHT,HHH]

. These are the possible outcomes of 3 tosses that guarantee a win.

Probability of winning is

\frac{[ |A| ]}{[ |\Omega| ]}

. The numbers of sample points of the event A divided by omega. There are 4 such sample points in A, so the probability is 4/8=0.5

Now suppose the coin is tossed once and we observe a head. That is to say event

B=[HTT,HTH,HHT,HHH]

has occured. This is the sample space of all possible future outcomes if we got a head on our first toss. So we need now another head to win the game. Our new sample space is B. Our event A is now just