Question #81791

According to the historical data gathered by an insurance company, the probability of a fire is 0.01. A manufacturer of fire alarms has determined through testing that a certain fire alarm has a 0.1 conditional probability to go off given that there is no fire, and a 0.2 conditional probability to remain silent given that there is a fire. The meaning of “go off” is that a fire alarm makes a loud sound or a bomb explodes.
Event A is that a fire breaks out.
Event B is that the fire alarm goes off.
(a) If you hear this fire alarm go off, what is the conditional probability that there is a fire? In
other words, what is P(A|B)?
(b) What is the conjunction probability, P(Bc ∩ Ac) ?

Expert's answer

Answer on Question #81791 – Math – Statistics and Probability

Question

According to the historical data gathered by an insurance company, the probability of a fire is 0.01. A manufacturer of fire alarms has determined through testing that a certain fire alarm has a 0.1 conditional probability to go off given that there is no fire, and a 0.2 conditional probability to remain silent given that there is a fire. The meaning of “go off” is that a fire alarm makes a loud sound or a bomb explodes.

Event A is that a fire breaks out. Event B is that the fire alarm goes off.

(a) If you hear this fire alarm go off, what is the conditional probability that there is a fire? In other words, what is P(A∣B)P(A|B)?

(b) What is the conjunction probability, P(BC∩AC)P(B^{C} \cap A^{C})?

Solution

(a) P(A)=0.01P(A) = 0.01

P(B∣AC)=0.1,P(BC∣A)=0.2P(B|A^C) = 0.1, P(B^C|A) = 0.2

According to Bayes’ rule


P(A∣B)=P(B∣A)P(A)P(B)P(A|B) = \frac{P(B|A)P(A)}{P(B)}


Apply the total probability theorem


P(B)=P(B∣A)P(A)+P(B∣AC)P(AC)P(B) = P(B|A)P(A) + P(B|A^C)P(A^C)P(B)=(1−0.2)(0.01)+0.1(1−0.01)=0.107P(B) = (1 - 0.2)(0.01) + 0.1(1 - 0.01) = 0.107


Hence,


P(A∣B)=(1−0.2)(0.01)0.107=8107P(A|B) = \frac{(1 - 0.2)(0.01)}{0.107} = \frac{8}{107}


(b) Conditional probability


P(BC∣AC)=P(BC∩AC)P(AC)P(B^C|A^C) = \frac{P(B^C \cap A^C)}{P(A^C)}P(BC∩AC)=P(BC∣AC)P(AC)P(B^C \cap A^C) = P(B^C|A^C)P(A^C)P(BC∩AC)=(1−0.1)(1−0.01)=0.891P(B^C \cap A^C) = (1 - 0.1)(1 - 0.01) = 0.891


Answer: a) P(A∣B)=8107P(A|B) = \frac{8}{107}; b) P(BC∩AC)=0.891P(B^C \cap A^C) = 0.891.

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