Probability of Detection 
There exist radars used for detecting aircrafts passing through the state of Iowa.  The state only has three radars, with each having a 25% chance of failing to detect a plane in the state.  Suddenly, a plane has entered the state.
   What is the probability the plane is detected?
  Let’s say the plane was detected.  What is the probability that at least two of the radars detected the plane?

 For the first question I computed $(1-.25)^3$ since there are three aircrafts and it would be a $75%$ percent chance the planes are detected.
 For the second part I am less sure but would it be $(.25)(.75)^2$ since there are two radars that detect and one that doesn’t?
 A: Part A: plane is not detected by all three radars with probability 
$$
\mathbb P(A^c)=0.25^3
$$
The probability that plane is detected (by at least one radar) is
$$
\mathbb P(A)=1-0.25^3.
$$
Part B: you have to find conditional probability that two radard detect plane (event $B$) if it is given that the plane is detected. It is
$$
\mathbb P(B\mid A) = \frac{\mathbb P(A\cap B)}{\mathbb P(A)}=\frac{\mathbb P(B)}{\mathbb P(A)} =\frac{\binom{3}{2}\cdot(1-0.25)^2\cdot 0.25}{1-0.25^3}=\frac{3\cdot(1-0.25)^2\cdot 0.25}{1-0.25^3}.
$$
Here $\mathbb P(A\cap B)=\mathbb P(B)$ since $B\subset A$.
Edit after the question is corrected 
If $C$ is an event that at least two radars detect plane, then 
$$
\mathbb P(C) = 3\cdot(1-0.25)^2\cdot 0.25 + (1-0.25)^3 
$$
and 
$$
\mathbb P(C\mid A) = \frac{\mathbb P(A\cap C)}{\mathbb P(A)}=\frac{\mathbb P(C)}{\mathbb P(A)}. 
$$
A: Part A:  $P_D =1-(.25)^3.$
Part B:  
$$P(\textrm{Exactly two detected})=3(.25)(.75)^2,$$
$$P(\textrm{Exactly three detected})=(.75)^3$$
so
$$P(\textrm{At least two detected}) = 3(.25)(.75)^2 + (.75)^3$$ 
and $$P(\textrm{Detected}|\textrm{At least two detected})=1,$$
we have
$$P(\textrm{At least two detected}|\textrm{Detected})=\frac{P(\textrm{Detected}|\textrm{At least two detected}) P(\textrm{At least two detected})}{P(\textrm{Detected})}$$
$$=\frac{ 3(.25)(.75)^2 +(.75)^3}{1-(.25)^3}.$$
A: For the first question you calculated the probability that all $3$ radars detected the plane. 
The correct answer is: $$1-0.25^3$$i.e. $1$ minus the probability that none of the $3$ radars detected the plane.

The second question is ambiguous.
If in the second question the words "two of the radars" is to be read as "exactly two of the radars" then you should find: $$P(R=2\mid R\neq0)$$ where $R$ has binomial distribution with parameters $n=3$ and $p=0.75$.
(Here $R$ stands for the number of radars that detected the plane)
If however the words must be read as "at least two of the radars" then you must go for finding: $$P(R\geq2\mid R\neq0)$$
The probabilities can be calculated by applying the general rule: $$P(A\mid B)=P(A\cap B)/P(B)$$
Can you do that yourself?
