Problem 34 - A First Course in Probability

Question

The monthly worldwide average number of airplane crashes of commercial airlines is \(2.2 .\) What is the probability that there will be (a) more than 2 such accidents in the next month? (b) more than 4 such accidents in the next 2 months? (c) more than 5 such accidents in the next 3 months? Explain your reasoning!

Step-by-Step Solution

Answer

The probability of having more than 2 accidents in the next month is \(P(X > 2) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) \right]\). Similarly, the probability of having more than 4 accidents in the next 2 months is \(P(X > 4) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right]\), and the probability of having more than 5 accidents in the next 3 months is \(P(X > 5) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \right]\). We use the Poisson distribution formula to calculate the cumulative probabilities and find the complement probabilities in each scenario.

Step 1: Part (a): More than 2 accidents in the next month

Step 1: Identify the parameters and set up the formula For this problem, k = 2, and λ = 2.2. The probability of more than 2 accidents in the next month is given by: \(P(X > 2) = 1 - P(X \leq 2)\) Step 2: Calculate the cumulative probabilities for k = 0, 1, and 2 Using the Poisson distribution formula: \(P(X = 0) = \frac{e^{-2.2} \cdot 2.2^0}{0!} \) \(P(X = 1) = \frac{e^{-2.2} \cdot 2.2^1}{1!}\) \(P(X = 2) = \frac{e^{-2.2} \cdot 2.2^2}{2!}\) Step 3: Calculate the complement probability The sum of these probabilities represents the probability of having at most 2 accidents in the next month. \(P(X > 2) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) \right]\) Finally, compute the value to get the probability of more than 2 accidents in the next month.

Step 2: Part (b): More than 4 accidents in the next 2 months

Step 1: Identify the parameters and set up the formula For this problem, k = 4, and since the time frame is 2 months, λ = 2.2 * 2 = 4.4. The probability of more than 4 accidents in the next 2 months is given by: \(P(X > 4) = 1 - P(X \leq 4)\) Step 2: Calculate the cumulative probabilities for k = 0, 1, 2, 3, and 4 Using the Poisson distribution formula and the updated λ value: \(P(X = 0) = \frac{e^{-4.4} \cdot 4.4^0}{0!} \) . . . \(P(X = 4) = \frac{e^{-4.4} \cdot 4.4^4}{4!}\) Step 3: Calculate the complement probability \(P(X > 4) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \right]\) Finally, compute the value to get the probability of more than 4 accidents in the next 2 months.

Step 3: Part (c): More than 5 accidents in the next 3 months

Step 1: Identify the parameters and set up the formula For this problem, k = 5, and since the time frame is 3 months, λ = 2.2 * 3 = 6.6. The probability of more than 5 accidents in the next 3 months is given by: \(P(X > 5) = 1 - P(X \leq 5)\) Step 2: Calculate the cumulative probabilities for k = 0, 1, 2, 3, 4, and 5 Using the Poisson distribution formula and the updated λ value: \(P(X = 0) = \frac{e^{-6.6} \cdot 6.6^0}{0!} \) . . . \(P(X = 5) = \frac{e^{-6.6} \cdot 6.6^5}{5!}\) Step 3: Calculate the complement probability \(P(X > 5) = 1 - \left[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \right]\) Finally, compute the value to get the probability of more than 5 accidents in the next 3 months.

Key Concepts

Poisson Distribution

The Poisson distribution is a probability model that describes the occurrence of rare events, or events that happen at a low, consistent rate within a fixed period or space. It's particularly useful for estimating the likelihood of a certain number of events happening over a specific interval of time or in a given space when these events occur independently of each other.

Mathematically, the Poisson distribution is expressed using the formula:
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
where:

• \( X \) is the random variable representing the number of events,
• \( k \) is the actual number of events that occur,
• \( \lambda \) is the average rate of occurrence (mean) of the event,
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

The mean rate \( \lambda \) can be thought of as the expected number of occurrences during the observed interval. In our exercise, \( \lambda \) changes according to the timeframe in question, allowing the Poisson model to adapt to different conditions.

Cumulative Probability

Cumulative probability refers to the probability that the variable takes a value less than or equal to a certain point. In the context of the Poisson distribution, it's the sum of probabilities for all outcomes up to a certain number and is extremely helpful in calculating the likelihood of multiple scenarios.

For example, to find the cumulative probability of \( k \) or fewer events in a Poisson distribution, you'd sum up the probabilities of there being 0 events, 1 event, up to and including \( k \) events:
\[ P(X \leq k) = \sum_{i=0}^{k} P(X = i) \]
This concept is key to understanding the remaining two exercises from the problem statement (parts b and c). These parts focus on more involved scenarios over extended periods requiring the computation of cumulative probabilities for a larger range of possible event counts.

Complement of a Probability

The complement of a probability is a foundational concept in probability theory, which describes the likelihood of an event not occurring. It is based on the principle that the total probability of all possible outcomes in a probability space is 1. Therefore, the probability of the complement is simply 1 minus the probability of the event occurring.

Mathematically, the complement of an event \( A \) is denoted as \( A' \) and the probability of the complement is given by:
\[ P(A') = 1 - P(A) \]
This concept is illustrated in parts (a), (b), and (c) of the exercise. To find the probability of having more than a certain number of events, we calculate the complement of the cumulative probability up to that number. This approach can be extremely useful in situations where directly calculating the desired probability is more complex than calculating the probability of all other alternatives and subtracting from 1.