A
C
D
E
F
B should be the obvious no-no
Answer:
A) 0.005
B) 0.001
C)0.0495
Step-by-step explanation:
Let A be the event that an aircraft is present and let B be the event the radar registers its presence.. Thus;
P(A) = Probability that an aircraft is present
P(A') = Probability that an aircraft is not present
P(B) = Probability that the radar generates an alarm
P(B') = Probability that the radar doesn't generate an alarm
Thus from what we are given, we have;
P(A) = 0.05
P(A') = 0.95
P(B) = 0.99
P(B') = 0.01
P(B|A') = 0.1
A) Probability of a false alarm will be;
P(A' ∩ B) = P(A) × P(B|A')
P(A' ∩ B) = 0.05 × 0.1 = 0.005
B) probability of missed detection is;
0.1 × (1 - 0.99) = 0.001
C) probability that an aircraft is present given that the radar registers a presence will be;
P(A ∩ B) = P(A) × P(B)
P(A ∩ B) = 0.05 × 0.99
P(A ∩ B) = 0.0495
You have to do like draw a number line and do the opposite and you'll get your answer and add them together and you'll get 45.5
Quite simple, actually...
so...just
simplify all of these numbers...
7.25%=0.0725
7.25*10(-3)--->>7.25*-30....
=-217.5
7 1/4..it's not that hard to change this one.. the

is really 0.25, or...
7.25
and finally,
0.725
So, we have...
0.0725,-217.5,7.25 and 0.725
Now, looking at them, they are really easy to see least to greatest...
-217.5, 0.0725, 0.725, and 7.25.
Thus, your answer.