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:

Let's define the events:

• \( B \): Break - in occurs ( \( P(B)=0.07 \) or \( 7\% \))
• \( \overline{B} \): No break - in occurs ( \( P(\overline{B}) = 1 - 0.07=0.93 \))
• \( A \): Alarm goes off

We know the conditional probabilities:

• \( P(A|B) \): 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 there was no break - in, \( P(A|\overline{B})=0.03 \)

We want to find \( P(B|A) \), the probability of a break - in given that the alarm goes off. We use Bayes' theorem:

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

Step 1: Calculate the numerator \( P(A|B)P(B) \)

Substitute \( P(A|B) = 0.98 \) and \( P(B)=0.07 \) into the formula:
\( P(A|B)P(B)=0.98\times0.07 = 0.0686 \)

Step 2: Calculate the denominator \( P(A|B)P(B)+P(A|\overline{B})P(\overline{B}) \)

First, calculate \( P(A|\overline{B})P(\overline{B}) \). Substitute \( P(A|\overline{B}) = 0.03 \) and \( P(\overline{B})=0.93 \):
\( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \)

Then, add the result from Step 1:
\( P(A|B)P(B)+P(A|\overline{B})P(\overline{B})=0.0686 + 0.0279=0.0965 \)

Step 3: Calculate \( P(B|A) \)

Divide the numerator by the denominator:
\( P(B|A)=\frac{0.0686}{0.0965}\approx0.7109 \) (Wait, there is a miscalculation above. Let's recalculate the denominator correctly. \( P(A|\overline{B}) = 0.03 \), \( P(\overline{B})=0.93 \), so \( P(A|\overline{B})P(\overline{B})=0.03\times0.93 = 0.0279 \). \( P(A|B)P(B)=0.98\times0.07 = 0.0686 \). Then \( 0.0686+0.0279 = 0.0965 \)? No, wait, 0.980.07 = 0.0686, 0.030.93=0.0279, sum is 0.0686 + 0.0279=0.0965? But 0.0686/0.0965≈0.7109? Wait, no, let's check the problem statement again. Wait, the problem says "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 \( P(A|B)=1 - 0.02 = 0.98 \), \( P(A|\overline{B}) = 0.03 \), \( P(B)=0.07 \), \( P(\overline{B}) = 0.93 \)

Bayes' theorem: \( P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|\overline{B})P(\overline{B})}=\frac{0.98\times0.07}{0.98\times0.07 + 0.03\times0.93}=\frac{0.0686}{0.0686+0.0279}=\frac{0.0686}{0.0965}\approx0.7109 \)? Wait, but let's recalculate 0.030.93: 0.030.93 = 0.0279, 0.98*0.07 = 0.0686, sum is 0.0686+0.0279 = 0.0965, 0.0686/0.0965≈0.7109. But wait, maybe I made a mistake in the problem understanding. Wait, the break - in rate is 7% (0.07), alarm off when break - in: 2%, so alarm on when break - in: 98% (0.98). Alarm on when no break - in: 3% (0.03). So the calculation is correct.

Wait, but let's check the options. Option e is 0.7109. But wait, let's re - do the calculation:

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

\( P(A|B)=1 - 0.02 = 0.98 \)

\( P(A|\overline{B}) = 0.03 \)

\( P(A)=P(A|B)P(B)+P(A|\overline{B})P(\overline{B})=0.98\times0.07 + 0.03\times0.93=0.0686+0.0279 = 0.0965 \)

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

Answer:

\( \boldsymbol{0.7109} \) (corresponding to option e)