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A parking lot has two entrances. Cars arrive at entrance I according to a Poisson distribution at an average of three per hour and at entrance II according to a Poisson distribution at an average of four per hour. What is the probability that a total of three cars will arrive at the parking lot in a given hour? (Assume that the numbers of cars arriving at the two entrances are independent.)

Short Answer

Expert verified
The probability that a total of three cars will arrive is approximately 0.05954.

Step by step solution

01

Understanding the Problem

We need to calculate the probability of having a total of three cars arriving at both entrances in a given hour. The arrivals at each entrance follow independent Poisson distributions: Entrance I has a rate of 3 cars per hour, and Entrance II has a rate of 4 cars per hour.
02

Combine the Poisson Parameters

When two independent Poisson distributions are combined, the result is also a Poisson distribution. We add the average rates for each entrance. Thus, the total average rate \( \lambda_t \) is \(3 + 4 = 7\) per hour.
03

Use the Poisson Formula

The probability of observing \(k\) events in a Poisson distribution with parameter \( \lambda \) is given by \(P(K=k) = \frac{e^{-\lambda} \lambda^k}{k!}\). Here, \(\lambda = 7\) and we want the probability of \(k = 3\) cars arriving.
04

Calculate the Probability

Substitute \(\lambda = 7 \) and \(k = 3 \) into the Poisson formula:\[P(K=3) = \frac{e^{-7} \, 7^3}{3!}\]Calculate \(e^{-7} \approx 0.000911882\) and \(7^3 = 343\). Thus:\[P(K=3) = \frac{0.000911882 \, \times \, 343}{6} = 0.05954\]Thus, the probability is approximately \(0.05954\).

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Key Concepts

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

Probability Calculation
Calculating the probability of an event is a core aspect of understanding how various outcomes occur. When analyzing the arrival of cars at a parking lot, the Poisson distribution is an appropriate model because it deals with the occurrence of independent events over a continuous interval. Poisson distribution helps us determine the likelihood of a given number of arrivals in a fixed period, given a known average rate.
  • The Poisson distribution is defined by the rate parameter \( \lambda \), representing the average number of occurrences in a given time frame.
  • The probability of observing exactly \( k \) events is derived using the formula \( P(K=k) = \frac{e^{-\lambda} \lambda^k}{k!} \).

When applying this to a real-world scenario, as in the parking lot problem, we're interested in how likely it is for an exact number of cars to arrive. By calculating the probability using the combined rate for both entrances, we leverage the properties of the Poisson distribution to get a precise probability measure.
Independent Events
In probability, understanding independent events is crucial for modeling complete systems accurately. Independent events are those whose outcomes do not influence each other. This concept is key to simplifying complex problems.
  • For example, the arrival of cars at entrance I and entrance II in our problem are independent events, meaning the occurrence at one entrance does not impact the other.
  • This independence allows us to sum the parameters \( \lambda_1 \) and \( \lambda_2 \) for the Poisson model at each entrance rather than considering their joint probabilities.

Hence, when combining the two separate Poisson distributions into a single one by adding their rates, we utilize the idea that independent Poisson processes can be merged without any complex interdependencies affecting their probabilities.
Combinatorial Properties
Combinatorial properties are handy in probability and statistics when determining how various outcomes can occur in a given setting. Although the problem might not explicitly mention combinatorics, it is underpinned by these principles when dealing with probabilities.
  • The Poisson formula itself involves combinatorial aspects, such as calculating \( k! \) (factorial of \( k \)), representing all possible combinations of events.
  • This computational aspect helps in understanding how many ways an event can occur, guiding the probability calculations.

When determining the probabilities like in our Poisson scenario, it's crucial to recognize how mathematical operations mirror combinatorial realities, making them vital for clear, concise probability computation.

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Most popular questions from this chapter

Suppose that \(Y\) is a binomial random variable based on \(n\) trials with success probability \(p\) and consider \(Y^{*}=n-Y\) a. Argue that for \(y^{*}=0,1, \ldots, n\) $$ P\left(Y^{*}=y^{*}\right)=P\left(n-Y=y^{*}\right)=P\left(Y=n-y^{*}\right) $$ b. Use the result from part (a) to show that $$ P\left(Y^{*}=y^{*}\right)=\left(\begin{array}{c} n \\ n-y^{*} \end{array}\right) p^{n-y^{*}} q^{y^{*}}=\left(\begin{array}{c} n \\ y^{*} \end{array}\right) q^{y^{*}} p^{n-y^{*}} $$ c. The result in part (b) implies that \(Y^{*}\) has a binomial distribution based on \(n\) trials and "success" probability \(p^{*}=q=1-p .\) Why is this result "obvious"?

Let \(Y\) denote a geometric random variable with probability of success \(p\) a. Show that for a positive integer \(a\) $$P(Y>a)=q^{a}$$, b. Show that for positive integers \(a\) and \(b\) ,$$P(Y>a+b | Y>a)=q^{b}=P(Y>b)$$ This result implies that, for example, \(P(Y>7 | Y>2)=P(Y>5) .\) Why do you think this property is called the memoryless property of the geometric distribution? c. In the development of the distribution of the geometric random variable, we assumed that the experiment consisted of conducting identical and independent trials until the first success was observed. In light of these assumptions, why is the result in part (b) "obvious"?

Given that we have already tossed a balanced coin ten times and obtained zero heads, what is the probability that we must toss it at least two more times to obtain the first head?

Of the volunteers donating blood in a clinic, \(80 \%\) have the Rhesus (Rh) factor present in their blood. a. If five volunteers are randomly selected, what is the probability that at least one does not have the Rh factor? b. If five volunteers are randomly selected, what is the probability that at most four have the Rh factor? c. What is the smallest number of volunteers who must be selected if we want to be at least \(90 \%\) certain that we obtain at least five donors with the Rh factor?

In May \(2005,\) Tony Blair was elected to an historic third term as the British prime minister. A Gallop U.K. poll (http://gallup.com/poll/content/default.aspx?ci=1710, June 28, 2005) conducted after Blair's election indicated that only \(32 \%\) of British adults would like to see their son or daughter grow up to become prime minister. If the same proportion of Americans would prefer that their son or daughter grow up to be president and 120 American adults are interviewed, a. what is the expected number of Americans who would prefer their child grow up to be president? b. what is the standard deviation of the number \(Y\) who would prefer that their child grow up to be president? c. is it likely that the number of Americans who prefer that their child grow up to be president exceeds 40?

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