Conditional Probability Testing Two tests are given sequentially on devices. 75% of the devices tested
pass the ﬁrst test. Also, 80% of the devices that passed the second test had also passed the ﬁrst one. Finally, 20% that failed the second test had passed the ﬁrst one.
(a) What is the probability that a given device passed the second test?
(b) Find the probability that, for a given device, the second test contradicts the ﬁrst one.
(c) Calculate the probability that a given device failed the second test, knowing that it passed the ﬁrst one.
My attempt:
Given:
$\Pr(A_p) = .25 $
$\Pr(A_p | B_p) = .80$
$\Pr(A_p | B_f) = .20 $
We must find I think:
(1) $\Pr(B_p) = \Pr(B_p ∩ A_p) + \Pr(B_p ∩ A_f) =\Pr(B_p | A_p)*\Pr(A_p) + \Pr(B_p | A_f)*\Pr(A_f)$
$=(.8)(.25)+??*(.75)$  
(2) ????
(3) $\Pr(B_f | A_p) =\frac{\Pr(B_f ∩ A_p)}{\Pr(A_p)} = \frac{\Pr(B_f)*\Pr(A_p | B_f)}{\Pr(A_p)} = \frac{(\text{answer to (1)})*(.20)}{(.75)}$
The Questions marks are were I need some help.
 A: Here's an answer to the first one. The others are very similar - basically all we're doing is playing around with identities and equalities a little. I'm going to rename things a little to make it easier to see why I'm doing things:
Your $A_p$ from now on will be called $A$, and your $B_p$ from now on will be called $B$. From this we can notice that $B_f$ will be $\overline{B}$. Hopefully this will make it a bit easier to see what's going on.
Let's start off using a few basic probability expressions.
$$ P(A)  = P(A \cap \overline{B}) + P(A \cap B)\\
P(A|B)P(B)=P(A\cap B) \\
P(A | \overline{B}) P(\overline{B})= P(A\cap \overline{B})$$
Add the second equation to the third:
$$P(A|B)P(B)+P(A | \overline{B}) P(\overline{B}) = P(A\cap B) +P(A\cap \overline{B}) $$
Remember $P(\overline{B})=1-P(B)$ and expand:
$$P(A|B)P(B)+P(A | \overline{B}) (1-P(B)) = P(A\cap B) +P(A\cap \overline{B})\\
P(A|B)P(B)+P(A|\overline{B})-P(A|\overline{B})P(B)=P(A\cap B) +P(A\cap \overline{B})$$
Now notice that the right side of our expression is the right side of the first identity I mentioned. So, let's substitute the other side of that identity in:
$$
P(A|B)P(B)+P(A|\overline{B})-P(A|\overline{B})P(B)=P(A)$$
And now notice that you have all of the probabilities in the equation above except for $P(B)$! So, let's plug in some numbers and solve for $P(B)$:
$$0.2-0.2P(B)+0.8P(B)=0.25 \\
P(B)=0.0833$$
Also, as a last note, for the second question you didn't write an inital expression so I'm going to assume you're not sure what it's asking for. It's saying, what is the probability that the first and second tests give different results - that would be the value for 
$$P(A\cap \overline{B})+P(\overline{A}\cap B)$$
