# Problem with equation of independent events

• Jun 19th 2007, 08:51 PM
computer-bot
Problem with equation of independent events
Please explain the question posted in the file attached.

CB
• Jun 19th 2007, 09:42 PM
CaptainBlack
Quote:

Originally Posted by computer-bot
Please explain the question posted in the file attached.

CB

$A \cap B$ is the set of simple events which are in both $A$ and $B$, and $|A \cap B|$ is the number of elements in this set.

It is the number of ways the both $A$ and $B$ can simultaneously occur.

RonL
• Jun 20th 2007, 06:24 AM
computer-bot
Quote:

Originally Posted by CaptainBlack
$A \cap B$ is the set of simple events which are in both $A$ and $B$, and $|A \cap B|$ is the number of elements in this set.

It is the number of ways the both $A$ and $B$ can simultaneously occur.

RonL

CaptainBlack I know what does A intersection B means but what I don't understand is that what does this statement |A and B|/|B| has to do with the occurence of two events. I mean the author said that "The probability that A occurs is P(A) = |A|/6 = 3/6 = 1/2, while presuming B occurs, the probability that A occurs is |A and B|/|B|. What I am trying to understand is that how did the author form this equation for this situation. I hope my question is clear.
• Jun 20th 2007, 08:08 AM
Plato
The probability of A happening given that B has occurred in symbols is
$P(A|B) = \frac{{P\left( {A \cap B} \right)}}{{P(B)}}$.
That explains the intersection in your question.

Thus if A and B are independent that means that
$P(A|B) = \frac{{P\left( {A \cap B} \right)}}{{P(B)}} = P(A)\quad \Rightarrow \quad P\left( {A \cap B} \right) = P(A)P(B)$.