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?
Best Answer
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 $$ \mathbb P(C\mid A) = \frac{\mathbb P(A\cap C)}{\mathbb P(A)}=\frac{\mathbb P(C)}{\mathbb P(A)}. $$