Are Events A and B Independent Based on Given Probabilities?

[frankfjf]

Homework Statement

Suppose P(A) = .6, P($$\overline{A}$$ | B) = .4. Check whether events A and B are independent.

Homework Equations

Two events A and B are said to be independent if P(A|B) = P(A). This is equivalent to stating that P(A $$\cap$$ B) = P(A)P(B)

If A and B are any two events, then the conditional probability of A given B, denoted by P(A | B), is P(A | B) = $$\frac{P(A \cap B)}{P(B)}$$ provided P(B) > 0.

The Attempt at a Solution

I know that P($$\overline{A}$$) = 1 - P(A) = .4

However, I'm not sure how to use the information given to check for independence. The professor says the solution should be very brief, but it's not coming to me.

Thinking on it a little more and looking it up online, it would seem that if P(A|B) = P(A) means A and B are independent, but P($$\overline{A}$$) != P(A), then A and B are dependent. But I'm not sure if this logic is correct.

 

[statdad]

You have P(A-complement | B) - it should be very simple to obtain P(A | B) from it.

Note: couldn't get the \bar{A} construct to work in Latex.

 

[frankfjf]

But that's my problem, how do I use that information to get P(A | B)?

 

[statdad]

But that's my problem, how do I use that information to get P(A | B)?

What do you get when you add P(A) to P(A-complement)? the same result works with P(A |B) and P(A-complement | B).

 

[frankfjf]

I get 1. So P(A | B) + P(A-complement | B) = P(A|B) + .4 = 1... P(A | B) = .6?