Problem 57 - A First Course in Probability

Question

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.

Step-by-Step Solution

Answer

\(P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)] = 0.5768\) \(P(X \geq 3 | X \geq 1) = \frac{P(X \geq 3)}{1 - P(X=0)} = \frac{0.5768}{1 - P(X=0)} = 0.6472\)

Step 1: Calculate the probability that 3 or more accidents occur today

We are looking for the probability P(X ≥ 3). To find this, we can instead find the complementary probability P(X < 3) and then subtract it from 1. So, \[P(X ≥ 3) = 1 - P(X < 3)\] which is equal to: \[P(X ≥ 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]\] Now, we use the Poisson PMF formula for each probability term in this sum: \[P(X=0) = \frac{e^{-3}(3)^0} {0!}\] \[P(X=1) = \frac{e^{-3}(3)^1} {1!}\] \[P(X=2) = \frac{e^{-3}(3)^2} {2!}\] Calculate each term and then find the sum: \[P(X ≥ 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]\]

Step 2: Calculate the probability that 3 or more accidents occur today, given that at least 1 accident occurs today

Now, let's find the conditional probability P(X ≥ 3 | X ≥ 1). We can use Bayes' theorem here: \[P(X ≥ 3 | X ≥ 1) = \frac{P(X ≥ 3 \cap X ≥ 1)}{P(X ≥ 1)}\] We know that if X ≥ 3, then X is also greater than or equal to 1. So, the intersection X ≥ 3 and X ≥ 1 is simply X ≥ 3. Therefore, \[P(X ≥ 3 | X ≥ 1) = \frac{P(X ≥ 3)}{P(X ≥ 1)}\] We already calculated P(X ≥ 3) in Step 1. Now we need to find P(X ≥ 1). This can be done by finding the complementary probability P(X < 1) and subtracting it from 1: \[P(X ≥ 1) = 1 - P(X=0)\] Using the Poisson PMF formula, we find P(X=0): \[P(X=0) = \frac{e^{-3}(3)^0} {0!}\] Now, we can substitute the values of P(X ≥ 3) and P(X = 0) into the equation of Step 2 to find the conditional probability: \[P(X ≥ 3 | X ≥ 1) = \frac{P(X ≥ 3)}{1 - P(X=0)}\]

Key Concepts

Probability Theory

Probability theory is a branch of mathematics focused on analyzing random phenomena. The core idea is to quantify the likelihood of different outcomes in experiments or events which are not deterministic. For example, when we toss a fair coin, the probability of getting heads is 0.5 because there are two equally likely outcomes.

Probability values range from 0 to 1, where 0 indicates an impossible event and 1 represents an event that is certain to occur. In the given exercise, we used probability theory to determine the likelihood of a certain number of accidents occurring on a highway. The uncertain nature of accidents makes probability theory an ideal tool to describe and analyze such events.

Poisson Distribution

The Poisson distribution is a probability distribution that expresses the probability of a given number of events occurring within a fixed interval of time or space, provided these events occur with a known constant rate and independently of the time since the last event.

It is defined by the parameter \(\lambda\), which represents the average number of events in the interval. The likelihood of observing exactly \(k\) events is given by the Poisson probability mass function (PMF):
\[P(X=k) = \frac{e^{-\lambda}(\lambda)^k}{k!}\]
The exercise asked for the probability of '3 or more' accidents which involves calculating the cumulative Poisson probabilities for 0, 1, and 2 accidents and then subtracting the sum from 1 to find the complementary probability.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. This concept is pivotal when dealing with interdependent events. The notation \(P(A|B)\) represents the probability of event \(A\) occurring, assuming that \(B\) has occurred.

In our context, the question in part (b) asked for the probability that there are '3 or more' accidents given that 'at least one' accident has already happened. This question transforms our analysis to a conditional framework, where the sample space is limited to days with at least one accident, hence altering the probabilities of subsequent accidents.

Bayes' Theorem

Bayes' theorem is a powerful formula used to update probabilities based on new information. It's expressed as:
\[P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}\]
Essentially, it relates the probability of \(A\) given \(B\) to the probability of \(B\) given \(A\). In the exercise, we applied a variation of Bayes’ theorem to find the conditional probability that there are '3 or more' accidents given that 'at least one' accident has occurred, by relating it to the unconditional probability of '3 or more' accidents.

To solve such problems, it's crucial to clearly define the events and use the correct probabilities, especially when handling conditional probabilities. This will lead to a correct application of Bayes' theorem and, therefore, the right solution.