Respuesta :
Answer:
The probability of a false alarm is 0.5116 .
Step-by-step explanation:
Let us indicate three events :
Event A = Alarm system triggers
Event B = Dangerous conditions exist
Event B' = Normal conditions exist
Now we are given with P(A/B) which means probability of alarm getting triggered given the dangerous conditions exist , P(A/B') which means probability of alarm getting triggered given the normal conditions exist and P(B) which means the probability of dangerous conditions existing i.e.
P(A/B) = 0.95 P(A/B') = 0.005 P(B) = 0.005 P(B') = 1 - P(B) = 0.995
(a) The probability of a false alarm means given the alarm gets triggered probability that there was normal conditions i.e. P(B'/A)
P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex]
Now P(A) = P(B) * P(A/B) + P(B') * P(A/B') {This representing Probability of alarm getting triggered in both the conditions]
P(A) = 0.005 * 0.95 + 0.995 * 0.005 = 9.725 x [tex]10^{-3}[/tex]
Since P(A/B') = 0.005
[tex]\frac{P(A\bigcap B')}{P(B')}[/tex] = 0.005 So, [tex]P(A\bigcap B')[/tex] = 0.005 * 0.995 = 4.975 x [tex]10^{-3}[/tex]
Therefore, P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex] = [tex]\frac{4.975*10^{-3} }{9.725*10^{-3} }[/tex] = 0.5116 .
