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Law Enforcement: Property Crime Does crime pay? The FBI Standard Survey of Crimes shows that for about \(80 \%\) of all property crimes (burglary, larceny, car theft, etc.), the criminals are never found and the case is never solved (Source: True Odds, by James Walsh, Merrit Publishing). Suppose a neighborhood district in a large city suffers repeated property crimes, not always perpetuated by the same criminals. The police are investigating six property crime cases in this district. (a) What is the probability that none of the crimes will ever be solved? (b) What is the probability that at least one crime will be solved? (c) What is the expected number of crimes that will be solved? What is the standard deviation? (d) Quota Problem How many property crimes \(n\) must the police investigate before they can be at least \(90 \%\) sure of solving one or more cases?

Step by step solution

01

Problem Interpretation

We are dealing with a binomial probability problem. We have six trials (property crimes), and the probability of success (solving a crime) for each trial is equal to 0.2, as only 20% of crimes are solved.

02

Probability of No Crimes Solved

The probability of solving none of the crimes can be found using the binomial formula: \( P(X = 0) = \binom{n}{0} (p)^0 (1-p)^n \), where \( n = 6 \), \( p = 0.2 \). This simplifies to \( P(X = 0) = (1-0.2)^6 \). Calculating this gives \( P(X = 0) = 0.8^6 \).

03

Calculate Probability of At Least One Solved

Use the complement rule to find the probability of at least one crime solved. This is \( 1 - P(X = 0) \). Using the result from the previous step, \( P(X \geq 1) = 1 - 0.8^6 \).

04

Expected Number of Solved Crimes

The expected value is given by \( E(X) = n \cdot p \) where \( n = 6 \) and \( p = 0.2 \). So, \( E(X) = 6 \times 0.2 \).

05

Calculate Standard Deviation

Standard deviation for a binomial distribution is given by \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \). Here, \( n=6 \), \( p=0.2 \). So, \( \sigma = \sqrt{6 \times 0.2 \times 0.8} \).

06

Determine Number of Crimes for 90% Success Probability

We want to find the smallest \( n \) such that \( 1 - (0.8)^n \geq 0.9 \). Solving for \( n \), we find \( n \geq \frac{\ln(0.1)}{\ln(0.8)} \). Calculating gives \( n \geq 10.97 \), so at least 11 crimes must be investigated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value

The expected value in a binomial distribution is an important concept that helps determine the average outcome of a random variable. This is particularly useful in scenarios like our crime investigation example, where understanding the average number of cases that might be solved could aid in resource allocation and planning.

The expected value for a binomial distribution can be calculated using the formula:

• \(E(X) = n \cdot p\)

where \(n\) is the number of trials (in this case, 6 property crimes), and \(p\) is the probability of success on a single trial (0.2 or 20%).
Calculating this gives:

• \(E(X) = 6 \times 0.2 = 1.2\)

This means that, on average, 1.2 property crimes are expected to be solved. While you can't solve "1.2 crimes" in reality, this statistical measure provides insight into the general effectiveness of crime-solving efforts in the district.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In a binomial distribution, it helps quantify how much the outcomes of various trials tend to deviate from the expected value.

For a binomial distribution, the standard deviation \(\sigma\) is calculated using:

• \(\sigma = \sqrt{n \cdot p \cdot (1-p)}\)

Substituting the values from our crime scenario:

• \(\sigma = \sqrt{6 \times 0.2 \times 0.8} \approx 0.98\)

This indicates that the number of crimes solved may typically vary by about 0.98 crimes from the average of 1.2. Understanding this variation is crucial for the police to gauge the unpredictability involved in solving crimes.

Complement Rule

The complement rule is a handy principle in probability that allows us to simplify calculations, particularly when we are interested in the occurrence of at least one event.

In our problem, if we wanted to calculate the probability of solving at least one crime, we first calculate the probability of not solving any crime and then use the complement rule:

• The probability of no crime being solved is \((1-p)^n\), where \(n = 6\) and \(p = 0.2\). This is:\[P(X=0) = 0.8^6\]
• Therefore, the probability of solving at least one crime is:\[P(X \geq 1) = 1 - 0.8^6\]

Using the complement rule makes finding such probabilities straightforward and less error-prone. This can be a particularly useful tool in cases where it is easier to state what doesn’t happen (no crimes solved) than what does (at least one solved).

Probability Calculation

Probability calculation within a binomial context involves determining the likelihood of a specific number of successes in a fixed number of trials. This is crucial for understanding potential outcomes and planning accordingly.

For the case of solving zero crimes, we use:

• \[P(X=0) = \binom{n}{0} (p)^0 (1-p)^n = (0.8)^6\]

Calculating this gives a probability of approximately 0.262.
To find the probability of solving at least one crime, we can use the complement rule:

• \[P(X \geq 1) = 1 - (0.8^6)\]

This results in a probability of about 0.738.
Understanding these calculations can help the police and neighborhood residents gauge their chances of case resolutions, thus aiding in realistic expectation setting and strategic planning.