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$$\cup$$B. 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$$\cup$$B$$^{c}$$) 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$$\cup$$B then A$$\cup$$B$$^{c}$$=A
Now, let x = P(A$$\cup$$B$$^{c}$$)=P(A)+P(B$$^{c}$$)-P(A$$\cap$$B$$^{c}$$)

Now if A and B$$^{c}$$ are independent, then their intersection is the same as multiplying them together. So P(A$$\cap$$B$$^{c}$$) = 0.8*0.5 = 0.4
This means that P(A$$\cup$$B$$^{c}$$) = 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

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