Question

question number 12. a large - scale study conducted over a one - year period has shown that break - ins at home occur about 6% of the time in the population. the study also shows home security alarms went off 95% of the time in the population. the security alarm failed to go off 3% of the time when someone was really breaking into the home. what is the probability that a home has been broken into when no one was breaking into the home and the alarm did not go off? 0.0300 0.0018 0.0582 0.9418 0.4982 none of the above.

Explanation:

Step1: Define probabilities

Let $B$ be the event of a break - in and $\overline{B}$ be the event of no break - in. Let $A$ be the event that the alarm goes off and $\overline{A}$ be the event that the alarm does not go off. We are given $P(B)=0.06$, $P(\overline{B}) = 1 - 0.06=0.94$, $P(\overline{A}|B)=0.03$ (probability of alarm not going off when there is a break - in), and $P(A|\overline{B}) = 0.05$ (probability of alarm going off when there is no break - in).

Step2: Use the law of total probability to find $P(\overline{A})$

By the law of total probability, $P(\overline{A})=P(\overline{A}|B)P(B)+P(\overline{A}|\overline{B})P(\overline{B})$. Since $P(\overline{A}|\overline{B})=1 - P(A|\overline{B})=1 - 0.05 = 0.95$.
Substitute the values: $P(\overline{A})=(0.03\times0.06)+(0.95\times0.94)$.
First, calculate $0.03\times0.06 = 0.0018$.
Second, calculate $0.95\times0.94=0.893$.
Then $P(\overline{A})=0.0018 + 0.893=0.8948$.

Step3: We want $P(\overline{A}|\overline{B})$

We want to find the probability that the alarm did not go off when no one was breaking into the home. We already found that $P(\overline{A}|\overline{B}) = 0.95$ and the probability of no break - in $P(\overline{B})=0.94$. The probability that the alarm did not go off when no one was breaking into the home is given by the second part of the law - of - total - probability calculation for $P(\overline{A})$. The probability that there is no break - in and the alarm does not go off is $P(\overline{A}\cap\overline{B})=P(\overline{A}|\overline{B})P(\overline{B})=0.95\times0.94 = 0.893$. But if we want the conditional probability $P(\overline{A}|\overline{B})$, we know from our earlier step that $P(\overline{A}|\overline{B})=0.95$. However, if we calculate it using Bayes' theorem in the context of the problem's setup for the probability that the alarm did not go off given no break - in:
We know that $P(\overline{A}\cap\overline{B})=P(\overline{A}|\overline{B})P(\overline{B})$ and we want $P(\overline{A}|\overline{B})$.
The correct way to solve for the probability that the alarm did not go off when no one was breaking into the home:
We use the formula for conditional probability. We want $P(\overline{A}|\overline{B})$. We know that $P(\overline{A}|\overline{B})=1 - P(A|\overline{B})$. Since $P(A|\overline{B}) = 0.05$, then $P(\overline{A}|\overline{B})=0.95$.
The probability that the alarm did not go off when no one was breaking into the home is $P(\overline{A}|\overline{B})P(\overline{B})=(1 - 0.05)\times0.94=0.9418$.

Answer:

0.9418