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Chapter 4: Problem 52

The monthly worldwide average number of airplane crashes of commercial airlines is \(3.5\). What is the probability that there will be (a) at least 2 such accidents in the next month; (b) at most 1 accident in the next month? Explain your reasoning!

Short Answer

Expert verified
(a) The probability of having at least 2 accidents in the next month is approximately 0.864, or 86.4%. (b) The probability of having at most 1 accident in the next month is approximately 0.1359, or 13.59%.

Step by step solution

01

Identify the distribution

Since the average number of events per interval is given, we can model the problem using a Poisson distribution. A Poisson distribution is represented by the formula: \[P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\] where: - \(P(X=k)\) is the probability of having \(k\) events in an interval - \(\lambda\) is the average number of events per interval (3.5 in this case) - \(e\) is the base of the natural logarithm (approximately 2.718) - and \(k\) is the number of events we are interested in. Now we can use this formula to find the probabilities for parts (a) and (b) of the exercise.
02

Calculate the probability for at least 2 events (part a)

We are asked to find the probability of having at least 2 accidents in the next month. This means we want to calculate the probability of having 2, 3, 4, etc. events. In other words, we need to calculate the complement of having 0 or 1 event: \[P(X\geq2) = 1 - P(X=0) - P(X=1)\] We can use the Poisson distribution formula to calculate the probabilities for \(X=0\) and \(X=1\): \[P(X=0) = \frac{e^{-3.5} 3.5^0}{0!} \approx 0.0302\] \[P(X=1) = \frac{e^{-3.5} 3.5^1}{1!} \approx 0.1057\] Now, plug in the values calculated back into the formula for at least 2 events: \[P(X\geq2) = 1 - 0.0302 - 0.1057 = 1 - 0.1359 \approx 0.864\] So, the probability of having at least 2 accidents in the next month is approximately 0.864, or 86.4%.
03

Calculate the probability for at most 1 event (part b)

We are asked to find the probability of having at most 1 accident in the next month. This means we want to calculate the probability of having 0 or 1 events: \[P(X\leq1) = P(X=0) + P(X=1)\] We already calculated the probabilities for \(X=0\) and \(X=1\) in Step 2: \[P(X\leq1) = 0.0302 + 0.1057 \approx 0.1359\] So, the probability of having at most 1 accident in the next month is approximately 0.1359, or 13.59%. To summarize the results: - The probability of having at least 2 accidents in the next month is approximately 0.864, or 86.4%. - The probability of having at most 1 accident in the next month is approximately 0.1359, or 13.59%.

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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 field of mathematics that deals with the likelihood of events occurring. It helps us make sense of uncertainty.
In the context of airplane crashes, we use probability to predict how often these rare events might happen. Probability is expressed as a number between 0 and 1, where 0 means the event will not occur and 1 means it certainly will.
  • In our problem, we are looking at whether there will be 2 or more crashes, or at most 1 crash, in a given month.
  • The calculation of probabilities helps us understand the chances of different outcomes based on historical data.
The Poisson distribution, a key concept here, models events happening at a constant mean rate within a fixed interval. It is particularly useful for rare, independent events occurring over time, like airplane crashes.
Expected Value
Expected value is a fundamental concept in probability theory, reflecting the average outcome if an experiment is repeated many times. It is calculated by summing the products of each possible outcome and its probability.
In the case of airplane crashes, the expected number of crashes per month is 3.5. This is represented by the symbol \(\lambda\) in a Poisson distribution.
  • The expected value doesn鈥檛 mean that exactly 3.5 crashes will happen each month, as it鈥檚 not possible to have a half-crash.
  • Instead, it tells us that over a long period, the average number of crashes per month will approach this number.
Recognizing the expected value helps us to understand the behavior of the distribution and the tendency of the observed data over time.
Complementary Probability
Complementary probability helps us see the relationship between the probability of an event occurring and not occurring. These complementary events together have probabilities that sum to 1.
In our airplane crash example, when calculating the probability of at least 2 crashes, we use complementary probability to simplify calculations by finding the probability of 0 or 1 crashes, and then subtracting from 1.
This approach is framed by:
  • The probability of at least 2 crashes is the complement of the probability of having 0 or 1 crash: \[P(X \geq 2) = 1 - P(X = 0) - P(X = 1)\]
  • Thus, complementary probability simplifies the calculation for cases where direct probability calculations are cumbersome.
Using complementary probability is powerful, as it often provides a quicker method to determine the likelihood of events in complex situations.

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Most popular questions from this chapter

A total of 4 buses carrying 148 students from the same school arrives at a football stadium. The buses carry, respectively, \(40,33,25\), and 50 students. One of the students is randomly selected. Let \(X\) denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let \(Y\) denote the number of students on her bus. (a) Which of \(E[X]\) or \(E[Y]\) do you think is larger? Why? (b) Comnute \(F[X]\) and \(E[Y)\)

Compare the Poisson approximation with the correct binomial probability for the following cases: (a) \(P\\{X=2\\}\) when \(n=8, p=.1\); (b) \(P\\{X=9\\}\) when \(n=10, p=.95\); (c) \(P\\{X=0\\}\) when \(n=10, p=.1\); (d) \(P\\{X=4\\}\) when \(n=9, p=.2\).

There are two possible causes for a breakdown of a machine. To check the first possibility. would cost \(C_{1}\) dollars, and, if that were the cause of the breakdown, the trouble could be repaired at a cost of \(R_{1}\) dollars. Similarly, there are costs \(C_{2}\) and \(R_{2}\) associated with the second possibility. Let \(p\) and \(1-p\) denote, respectively, the probabilities that the breakdown is caused by the first and second possibilities. Under what conditions on \(p, C_{i}, R_{i}\), \(i=1,2\), should we check the first possible cause of breakdown and then the second, as opposed to reversing the checking order, so as to minimize the expected cost involved in returning the machine to working order? NOTE: If the first check is negative, we must still check the other possibility.

If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is \(\frac{1}{100}\), what is the (approximate) probability that you will win a prize (a) at least once; (b) exactly once; (c) at least twice?

Suppose that two teams play a series of games that ends when one of them has won \(i\) games. Suppose that each game played is, independently, won by player \(A\) with probability \(p\). Find the expected number of games that are played when (a) \(i=2\) and (b) \(i=3\). Also show in both cases that this number is maximized when \(p=\frac{1}{2}\).
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