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Chapter 4: Problem 57

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter \(\lambda=3\) (a) Find the probability that 3 or more accidents occur today. (b) Repeat part (a) under the assumption that at least 1 accident occurs today.

Short Answer

Expert verified
\(P(X \geq 3) = 1 - (e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2})\) \(P(X \geq 3 | X \geq 1) = \frac{1 - (e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2})}{1 - e^{-3}}\)

Step by step solution

01

Calculate Probability of 3 or More Accidents

First, we will calculate the probability that 3 or more accidents occur today. Given a Poisson random variable \(X\) with parameter \(\lambda = 3\), the probability mass function (PMF) is given by \[P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!},\] where \(k\) is the number of accidents. Since we want the probability that 3 or more accidents occur today, we take the complement which is easier to calculate: the probability of having 0, 1, or 2 accidents. \(P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X=0) + P(X=1) + P(X=2))\) Let's calculate these probabilities using the PMF: \(P(X=0) = \frac{e^{-3}3^0}{0!} = e^{-3}\) \(P(X=1) = \frac{e^{-3}3^1}{1!} = 3e^{-3}\) \(P(X=2) = \frac{e^{-3}3^2}{2!} = \frac{9e^{-3}}{2}\) Now, we substitute the probabilities back into the equation: \(P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2})\)
02

Calculate Conditional Probability of 3 or More Accidents Given at Least 1 Accident

Now, we want to calculate the probability of having 3 or more accidents today, given that at least 1 accident occurs today. The conditional probability can be calculated using the formula: \(P(X \geq 3 | X \geq 1) = \frac{P((X \geq 3) \cap (X \geq 1))}{P(X \geq 1)}\) Since \(X\geq 3\) already implies \(X\geq 1\), the numerator will be \(P(X\geq 3)\). For the denominator, we simply need to find the probability of at least 1 accident. \(P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-3}\) Substitute these probabilities into the conditional probability formula: \(P(X \geq 3 | X \geq 1) = \frac{P(X \geq 3)}{P(X \geq 1)} = \frac{1 - (e^{-3} + 3e^{-3} + \frac{9e^{-3}}{2})}{1 - e^{-3}}\) We have successfully found the probabilities of both (a) and (b) parts.

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Key Concepts

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

Probability Mass Function (PMF)
Understanding the probability mass function (PMF) is crucial when dealing with discrete random variables such as the Poisson random variable presented in the exercise. The PMF describes the probability that a random variable takes on a particular value. For a Poisson random variable with parameter \(\lambda\text{, the PMF is given as}:\)

\[P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!},\]
where \(e\) is the base of the natural logarithm, \(k\) is the specific number of occurrences, and \(k!\) denotes \(k\) factorial. To compute the probability of observing exactly \(k\) events, you use the PMF formula by plugging in the value of \(k\) and the known rate \(\lambda\). For example, if \(\lambda = 3\), to find the probability of zero accidents, we calculate \(P(X=0)\) using the PMF with \(k=0\), which results in \(e^{-3}\).
Using the PMF helps us understand the likelihood of different outcomes and is the starting point for many other calculations in probability.
Conditional Probability
When we're interested in the probability of an event given that another event has already occurred, we're in the realm of conditional probability. It aids in refining our predictions based on new information. In the example from the exercise, we want to find the probability of having 3 or more accidents, given that at least one accident has already occurred. Mathematically, this is expressed as:

\[P(X \geq 3 | X \geq 1)\]
In conditional probability, the \(|\) symbol reads as 'given that'. To calculate it, we divide the probability of both events happening by the probability of the given condition:

\[P(A|B) = \frac{P(A \cap B)}{P(B)},\]
where \(P(A|B)\) is the conditional probability of event \(A\) given \(B\), \(P(A \cap B)\) represents the probability of both \(A\) and \(B\) occurring together, and \(P(B)\) is the probability of the given condition \(B\). In practice, such as in our traffic accident example, this could be used to adjust safety measures on days when the baseline risk is already known to be higher.
Complement Rule
The complement rule is a foundational concept in probability that relates the probability of an event to the probability of its complement. The complement of an event \(A\) consists of all outcomes not in \(A\). Symbolically, the complement rule states that:

\[P(A) + P(A^c) = 1,\]
where \(P(A^c)\) represents the probability of the complement of event \(A\). This rule is particularly handy when the probability of the complement is easier to calculate than the probability of the event itself.
For example, in our Poisson random variable scenario, calculating the probability of having 3 or more accidents (\(P(X \geq 3)\)) directly might be cumbersome due to the need to calculate many probabilities for \(k=3, k=4,\) and so on. Instead, the complement rule allows us to calculate the probability of having fewer than 3 accidents and subtract that from 1, providing a much simpler calculation path:

\[P(X \geq 3) = 1 - P(X < 3).\]
This approach is practical and widely used in solving probability problems, making the complement rule an invaluable tool in a probability toolkit.

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