Question

1. a large-scale study conducted over a one-year period has shown that break-ins at home occur about 7% of the time in the population. the study also shows home security alarms went off 3% of the time when no one was breaking into the home. the security alarm failed to go off 2% of the time when someone was really breaking into the home. if an alarm is going off, what is the probability that the house was broken into? a) 0.0686 b) .9800 c) .9314 d) .2109 e) .7109

Explanation:

Step1: Define Events

Let \( B \) be the event that a break - in occurs, \( \overline{B} \) be the event that no break - in occurs, \( A \) be the event that the alarm goes off.

We know:

• \( P(B)=0.07 \) (break - ins occur about 7% of the time)
• \( P(\overline{B}) = 1 - P(B)=1 - 0.07 = 0.93 \)
• \( P(A|B) \): Probability that alarm goes off given a break - in. Since the alarm failed to go off 2% of the time when there was a break - in, \( P(A|B)=1 - 0.02=0.98 \)
• \( P(A|\overline{B}) \): Probability that alarm goes off given no break - in. Since the alarm went off 3% of the time when no one was breaking in, \( P(A|\overline{B}) = 0.03 \)

Step2: Apply Bayes' Theorem

Bayes' theorem states that \( P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|\overline{B})P(\overline{B})} \)

First, calculate the numerator: \( P(A|B)P(B)=0.98\times0.07 = 0.0686 \)

Then, calculate the denominator:
\( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \)
\( P(A|B)P(B)+P(A|\overline{B})P(\overline{B})=0.0686 + 0.0279=0.0965 \)

Step3: Calculate \( P(B|A) \)

\( P(B|A)=\frac{0.0686}{0.0965}\approx0.7109 \) (Wait, no, wait. Wait, I made a mistake in the denominator calculation. Wait, \( P(A|\overline{B}) \) is the probability that alarm goes off when there is no break - in. Wait, the problem says "the home security alarms went off 3% of the time when no one was breaking into the home" so \( P(A|\overline{B}) = 0.03 \), and \( P(\overline{B})=0.93 \), so \( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \). And \( P(A|B)P(B)=0.98\times0.07 = 0.0686 \). Then the denominator is \( 0.0686+0.0279 = 0.0965 \)? Wait, no, wait, the user's options: let's re - check.

Wait, maybe I misread the problem. Let's re - read: "the home security alarms went off 3% of the time when no one was breaking into the home. The security alarm failed to go off 2% of the time when someone was really breaking into the home."

So \( P(A|\overline{B}) = 0.03 \) (alarm off when no break - in? No, wait: "went off 3% of the time when no one was breaking into the home" so \( P(A|\overline{B})=0.03 \). "failed to go off 2% of the time when someone was really breaking into the home" so \( P(\overline{A}|B)=0.02 \), so \( P(A|B)=1 - 0.02 = 0.98 \).

\( P(B) = 0.07 \), \( P(\overline{B})=0.93 \)

By Bayes' theorem:

\( P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|\overline{B})P(\overline{B})} \)

\( P(A|B)P(B)=0.98\times0.07 = 0.0686 \)

\( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \)

Denominator: \( 0.0686 + 0.0279=0.0965 \)

\( P(B|A)=\frac{0.0686}{0.0965}\approx0.7109 \)? Wait, but let's check the options. Option e is 0.7109. Wait, but maybe I made a mistake in the interpretation. Wait, maybe "failed to go off 2% of the time when someone was really breaking into the home" means \( P(\overline{A}|B) = 0.02 \), so \( P(A|B)=0.98 \), correct. "went off 3% of the time when no one was breaking into the home" means \( P(A|\overline{B}) = 0.03 \), correct.

Wait, let's recalculate:

\( P(A|B)P(B)=0.98\times0.07 = 0.0686 \)

\( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \)

Sum: \( 0.0686+0.0279 = 0.0965 \)

\( P(B|A)=\frac{0.0686}{0.0965}\approx0.7109 \)

Answer:

e).7109