91Ó°ÊÓ

Chances: Risk and Odds in Everyday Life, by James Burke, reports that the probability a police officer will be killed in the line of duty is 0.5\% (or less). (a) In a police precinct with 175 officers, let \(r=\) number of police officers killed in the line of duty. Explain why the Poisson approximation to the binomial would be a good choice for the random variable \(r .\) What is \(n ?\) What is \(p ?\) What is \(\lambda\) to the nearest tenth? (b) What is the probability that no officer in this precinct will be killed in the line of duty? (c) What is the probability that one or more officers in this precinct will be killed in the line of duty? (d) What is the probability that two or more officers in this precinct will be killed in the line of duty?

Step by step solution

01

Understanding Poisson Approximation

The Poisson approximation to the binomial distribution is suitable when the probability of an event is small (, and the number of trials is large (\(n\)). In this case, \(n = 175\) (the number of officers) and \(p = 0.005\) (the probability an officer will be killed on duty is 0.5%). We approximate the binomial distribution with a Poisson distribution because \(np\) (expected number) is small and \(n\) is relatively large.

02

Calculate Lambda (\(\lambda\))

For a Poisson distribution, \( \lambda = np \). Here, \( n = 175 \) and \( p = 0.005 \), so \( \lambda = 175 \times 0.005 = 0.875 \). Thus, \( \lambda \approx 0.9 \) to the nearest tenth.

03

Probability of Zero Officers Killed (\( P(r=0) \))

For a Poisson distribution, the probability that \( r \) events occur is given by the formula \( P(r=k) = \frac{e^{-\lambda} \lambda^k}{k!} \). To find \( P(r=0) \), substitute \( k = 0 \) and \( \lambda = 0.9 \). The probability is \( P(r=0) = \frac{e^{-0.9} \cdot 0.9^0}{0!} = e^{-0.9} \approx 0.4066 \).

04

Probability of One or More Officers Killed (\( P(r \geq 1) \))

To find the probability of one or more officers being killed, we use the complement rule: \( P(r \geq 1) = 1 - P(r=0) \). From the previous step, \( P(r=0) = 0.4066 \), thus, \( P(r \geq 1) = 1 - 0.4066 \approx 0.5934 \).

05

Probability of Two or More Officers Killed (\( P(r \geq 2) \))

The probability of two or more officers being killed is \( P(r \geq 2) = 1 - P(r = 0) - P(r = 1) \). First, calculate \( P(r=1) \): \( P(r=1) = \frac{e^{-0.9} \cdot 0.9^1}{1!} = 0.9 \cdot e^{-0.9} \approx 0.3659 \). Therefore, \( P(r \geq 2) = 1 - 0.4066 - 0.3659 = 0.2275 \).

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Key Concepts

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

Binomial Distribution

The binomial distribution is a probability distribution that summarizes the likelihood of a variable taking on one of two independent values under a specified set of parameters or assumptions.
Here are the key assumptions:

• There are fixed number of trials, denoted by \( n \).
• Each trial is independent.
• There are only two possible outcomes in each trial: success or failure.
• The probability of success, \( p \), is constant for each trial.

In our exercise, 175 police officers represent the trials, and each officer has a small probability of being killed (\( p = 0.005 \)).
Owing to the small probability and large number of trials, the binomial distribution is well-suited for approximation by a Poisson distribution, which simplifies calculations without losing accuracy.
The Poisson distribution is often used as it can handle situations with a large \( n \) and small \( p \) efficiently.

Probability

Probability is a branch of mathematics concerned with the likelihood of an event's occurrence, expressed between 0 and 1. A higher probability means a greater chance of the event occurring.
In our problem:

• The probability of no police officers being killed is \( P(r=0) \), computed using the Poisson formula.
• Additionally, probabilities such as one or more, or two or more officers being affected can be derived using these calculations and complement rules.

For Poisson distributions, probabilities are computed using:\[P(r=k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}\]Here, \( \lambda \) is the mean number of occurrences, and \( k \) is the number of events.
Calculating different probabilities using the above formula ensures you understand how likely each scenario is, aiding in decision-making and risk assessment in real-world situations.

Expected Value

The expected value is a fundamental concept in probability, representing the average or mean value of a random variable after numerous trials. It provides an estimation of what to expect in a given probabilistic situation.
For the binomial distribution, the expected value is given by \( E(X) = n \cdot p \).

• In our problem, with \( n = 175 \) and \( p = 0.005 \), the expected value is \( 0.875 \).
• This value tells us, on average, how many officers might be expected to be involved in such tragic events.

For the Poisson distribution approximation, the expected value equals \( \lambda \), which is also \( np \). This similarity is why the Poisson distribution effectively approximates the binomial distribution under certain conditions.
Understanding the expected value helps in grasping the conceptual pivot between theoretical distributions and actual expectations in various scenarios.