Alternative formula for the Rule of Addition not working for some problems

[Juwane]

In probability, the rule of addition is:

[tex]P( A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

This can be written as

[tex]P( A \cup B) = P(A) + P(B) - P(A)P(B)[/tex]

However, I've seen that using both formulas for the same problem gives different answers in some problems. Does anyone know why? Is there a condition for applying the second formula?

 

[CRGreathouse]

Is there a condition for applying the second formula?

Yes, the two must be independent. The first one works in all cases.

 

[Juwane]

But consider this example:

A shopper goes to a fruit store. The probability that he will buy (1) bananas is 0.40, (2) oranges is 0.30, and (3) both bananas and oranges is 0.2. What is the probability that the shopper will buy bananas, oranges, or both?


Using the usual formula:

[tex]P( B \cup O) = P(B) + P(O) - P(B \cap O) = 0.40 + 0.30 - 0.2 = 0.50[/tex]

Using the alternate formula:

[tex]P( B \cup O) = P(B) + P(O) - P(A)P(B) = 0.40 + 0.30 - (0.4)(0.3) = 0.58[/tex]

Why they are different?

 

[CRGreathouse]

Why they are different?

Because the probabilities are not independent. (Didn't you see my post?)

 

[Juwane]

Because the probabilities are not independent. (Didn't you see my post?)

I did see your post. That's why I said, "But consider this example".

Anyway, when the probabilities are independent, we don't need the minus part of the formula. This is enough:

[tex]P( A \cup B) = P(A) + P(B)[/tex]

Then what is the use of writing the formula in the alternate form?

 

[CRGreathouse]

That's disjoint, not independent.

 

[SW VandeCarr]

This is enough:

[tex]P( A \cup B) = P(A) + P(B)[/tex]

Then what is the use of writing the formula in the alternate form?

Only if probabilities are independent does [tex] P(A)P(B)=P(A)\cap P(B)[/tex]. Otherwise you must be given or otherwise determine the intersection.

EDIT: A zero value for the intersection of P(A) and P(B) is not consistent with independence for non zero P(A),P(B).