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Parents who are concerned that their children are "accident prone" can be reassured, according to a study conducted by the Department of Pediatrics at the University of California, San Francisco. Children who are injured two or more times tend to sustain these injuries during a relatively limited time, usually 1 year or less. If the average number of injuries per year for school- age children is two, what are the probabilities of these events? a. A child will sustain two injuries during the year. b. A child will sustain two or more injuries during the year. c. A child will sustain at most one injury during the year.

Short Answer

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Question: Based on the given average number of injuries per year for school-age children, what are the probabilities for: a. A child sustaining exactly two injuries during the year b. A child sustaining two or more injuries during the year c. A child sustaining at most one injury during the year Answer: a. The probability of a child sustaining exactly two injuries during the year is approximately 0.2707 or 27.07%. b. The probability of a child sustaining two or more injuries during the year is approximately 0.594 or 59.4%. c. The probability of a child sustaining at most one injury during the year is approximately 0.406 or 40.6%.

Step by step solution

01

Case a: A child will sustain exactly two injuries

We have k=2 and λ=2. We will use the Poisson probability formula to find the probability: P(X = 2) = (e^{-2} * 2^2) / 2! P(X = 2) ≈ (0.13534 * 4) / 2 ≈ 0.2707 The probability that a child will sustain exactly two injuries during the year is approximately 0.2707 or 27.07%.
02

Case b: A child will sustain two or more injuries

In this case, we need to find the probability that a child will sustain 2 or more injuries. We can find the probability by using the complement rule - that is, we find the probability of a child sustaining 1 injury or none and subtract it from 1. P(X >= 2) = 1 - [P(X = 0) + P(X = 1)] We will use the Poisson probability formula for k=0 and k=1: P(X = 0) = (e^{-2} * 2^0) / 0! ≈ 0.13534 P(X = 1) = (e^{-2} * 2^1) / 1! ≈ 0.2707 P(X >= 2) = 1 - [0.13534 + 0.2707] ≈ 1 - 0.406 = 0.594 The probability that a child will sustain two or more injuries during the year is approximately 0.594 or 59.4%.
03

Case c: A child will sustain at most one injury

In this case, we want to find the probability of a child sustaining 0 or 1 injury. We already computed the probabilities for k=0 and k=1 in the previous step. P(X ≤ 1) = P(X = 0) + P(X = 1) ≈ 0.13534 + 0.2707 ≈ 0.406 The probability that a child will sustain at most one injury during the year is approximately 0.406 or 40.6%.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random phenomena. The central objects of probability theory are random variables, which represent numerical values that occur randomly. One of its primary aims is to provide a mathematical foundation for making statements about the likelihood of various outcomes.

When we talk about the probability of something happening, we refer to how likely it is to occur, measured on a scale from 0 (impossible) to 1 (certain). For example, flipping a coin has two possible outcomes: heads or tails. If the coin is fair, each outcome has a probability of 0.5. Probability theory allows us to quantify and predict the frequency of events, such as the number of times a child might be injured in a year, based on a given set of parameters.

In the textbook exercise, we analyze the number of injuries a child might sustain in a year. With a given average, we can calculate the exact probabilities using specific probability distributions, which in this case is the Poisson distribution.
Poisson Probability Formula
The Poisson probability formula is used when we are interested in the number of times an event occurs in a fixed interval of time or space. The event should be random, and occurrences should be independent of each other. The formula is given by:
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
Where:\begin{itemize}\item \( P(X = k) \) is the probability of observing \( k \) events.\item \( e \) is the base of the natural logarithm (approximately equal to 2.71828).\item \( \lambda \) is the average number of occurrences in the interval (also known as the rate parameter).\item \( k \) is the actual number of occurrences for which the probability is being calculated.\item \( k! \) is the factorial of \( k \).\end{itemize}In the textbook scenario, the Poisson formula helps us calculate the probabilities of a child sustaining a specific number of injuries in a year, given the average number of injuries is two. Understanding this formula is crucial as it enables us to determine the probability of various occurrences and can be applied to numerous real-world situations where events happen at a constant average rate.
Complement Rule in Probability
The complement rule in probability is a fundamental concept that helps us find the probability of the occurrence of an event by subtracting the probability of the event not occurring from 1. The rule states that the sum of the probabilities of an event and its complement is always equal to 1 (probability of a sure event). In mathematical terms, for an event A, the compliment rule is expressed as:
\[ P(A) + P(A') = 1 \]
Where \( P(A) \) is the probability of the event, and \( P(A') \) is the probability of the complement of the event -- the event not occurring. To find \( P(A) \), you can subtract \( P(A') \) from 1:
\[ P(A) = 1 - P(A') \]
In the exercise, this rule allows us to calculate the probability of a child sustaining two or more injuries by subtracting the probability of sustaining no more than one injury from 1. This is a powerful concept because it sometimes makes calculations easier, especially when it is difficult to directly compute the probability of the occurrence of an event.

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