SOLUTION: Show that two events A and B are independent if and only if P(A ∩ B) = P(A)P(B) when at least one of P(A) or P(B) is not zero thank you :)

Algebra ->  Probability-and-statistics -> SOLUTION: Show that two events A and B are independent if and only if P(A ∩ B) = P(A)P(B) when at least one of P(A) or P(B) is not zero thank you :)      Log On


   



Question 1177192: Show that two events A and B are independent if and only if P(A ∩ B) = P(A)P(B) when
at least one of P(A) or P(B) is not zero
thank you :)

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down the proof of the independence of events A and B.
**Definition of Independence:**
Two events A and B are independent if the occurrence of one event does not affect the probability of the other. Mathematically, this is expressed as:
* P(A|B) = P(A) if P(B) > 0
* P(B|A) = P(B) if P(A) > 0
**Proof:**
We need to show that A and B are independent if and only if P(A ∩ B) = P(A)P(B).
**1. If A and B are independent, then P(A ∩ B) = P(A)P(B):**
* Assume A and B are independent.
* By the definition of conditional probability, we have:
* P(A|B) = P(A ∩ B) / P(B) (if P(B) > 0)
* P(B|A) = P(A ∩ B) / P(A) (if P(A) > 0)
* Since A and B are independent, we know that P(A|B) = P(A) (if P(B) > 0) and P(B|A)=P(B) (if P(A) > 0).
* Substituting P(A|B) = P(A) into the first equation:
* P(A) = P(A ∩ B) / P(B)
* P(A ∩ B) = P(A)P(B) (if P(B) > 0)
* Substituting P(B|A) = P(B) into the second equation:
* P(B) = P(A ∩ B) / P(A)
* P(A ∩ B) = P(A)P(B) (if P(A) > 0)
* Therefore, if A and B are independent, then P(A ∩ B) = P(A)P(B) when at least one of P(A) or P(B) is not zero.
**2. If P(A ∩ B) = P(A)P(B), then A and B are independent:**
* Assume P(A ∩ B) = P(A)P(B).
* We want to show that P(A|B) = P(A) (if P(B) > 0) and P(B|A)=P(B) (if P(A) > 0).
* Using the definition of conditional probability:
* P(A|B) = P(A ∩ B) / P(B) (if P(B) > 0)
* P(B|A) = P(A ∩ B) / P(A) (if P(A) > 0)
* Substituting P(A ∩ B) = P(A)P(B) into the first equation:
* P(A|B) = (P(A)P(B)) / P(B)
* P(A|B) = P(A) (if P(B) > 0)
* Substituting P(A ∩ B) = P(A)P(B) into the second equation:
* P(B|A) = (P(A)P(B)) / P(A)
* P(B|A) = P(B) (if P(A) > 0)
* Therefore, if P(A ∩ B) = P(A)P(B), then A and B are independent when at least one of P(A) or P(B) is not zero.
**Conclusion:**
We have shown that A and B are independent if and only if P(A ∩ B) = P(A)P(B) when at least one of P(A) or P(B) is not zero. This completes the proof.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

The problem's formulation is  INCORRECT.

The  DEFINITION  of independence of two events is  THIS:

        Two events,  A  and  B,  are called independent if and only if

                P(A intersection B) = P(A)*P(B)

        without any other restrictions on  P(A)  and/or  P(B).


As you brought this problem to the forum as it is in the post,
it means that you copy-pasted it from a bad source, which is anti-pedagogic.