Independent Probability Question

[tomtom690]

Homework Statement

Hello, I just want to know if I am going about this the right way.
A and B are outcomes of a random experiment in a sample space \Omega such that \Omega = A\cupB. P(A) = 0.8 and P(B) = 0.5 Study if A and B, A and B^{c}, A^{c} and B, and A^{c} and B^{c} are independent outcomes. Also, evaluate P(A\cupB^{c}) etc.

Homework Equations

The Attempt at a Solution

For the first three, I have used the same reasoning. I shall give an example of A and B^{c}.
Since \Omega = A\cupB then A\cupB^{c}=A
Now, let x = P(A\cupB^{c})=P(A)+P(B^{c})-P(A\capB^{c})

Now if A and B^{c} are independent, then their intersection is the same as multiplying them together. So P(A\capB^{c}) = 0.8*0.5 = 0.4
This means that P(A\cupB^{c}) = 0.9 \neq P(A) = 0.8, so they are not independent.

However I am having difficulty applying this reasoning (if correct!) to the final one. And then the evaluation part seems too easy, as I have already said in this working out what they are equal to.

Thanks.

 

[jbunniii]

P(A^c \cap B^c) = 1 - P(A \cup B) = 0

whereas

P(A^c)P(B^c) = (0.2)(0.5) \neq 0

so A^c and B^c are not independent.

 

[jbunniii]

As for the other questions, I calculated

P(A \cup B) = P(A) + P(B) - P(A \cap B)

so

1 = 0.8 + 0.5 - P(A \cap B)

which means that

P(A \cap B) = 0.3

Now use

P(A) = P(A \cap B) + P(A \cap B^c)

to establish that

P(A \cap B^c) = 0.5

which disagrees with your calculation.

[Edit] Oh wait, I see what you did. Yes, I think your way is OK, too.

If you use my calculation, you would observe that

P(A)P(B^c) = (0.8)(0.5) = 0.4

which does not equal P(A \cap B^c), so A and B^c are not independent.

You can then calculate

P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = 0.8