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According to a report by the American Academy of Orthopedic Surgeons, \(29 \%\) of pedestrians admit to texting while walking. Suppose two pedestrians are randomly selected. a. If the pedestrian texts while walking, record a \(\mathrm{T}\). If not, record an \(\mathrm{N}\). List all possible sequences of Ts and Ns for the two pedestrians. b. For each sequence, find the probability that it will occur by assuming independence. c. What is the probability that neither pedestrian texts while walking? d. What is the probability that both pedestrians text while walking? e. What is the probability that exactly one of the pedestrians texts while walking?

Short Answer

Expert verified
The possible sequences of Ts and Ns for the two pedestrians are TT, TN, NT and NN. The respective probabilities for these sequences are about 0.0841, 0.4118, 0.4118, and 0.5041 respectively. Probability that neither pedestrian texts while walking is about 0.5041. Probability that both pedestrians text while walking is about 0.0841. And, probability that exactly one pedestrian is texting while walking is about 0.4118.

Step by step solution

01

Identify the possibilities

Firstly, we need to figure out the possible sequences of Ts (Texting) and Ns (Not texting) for the two pedestrians selected randomly. Possible sequences are TT, TN, NT, and NN.
02

Assign probabilities to each possibility

Since the problem requires us to assume independence, we can use the given percentage of texting pedestrians (29%) to find the probability of each possible sequence. For each sequence, the probability would be \(0.29^x * 0.71^y\) where \(x\) is the number of Ts and \(y\) is the number of Ns. The probability for TT is \(0.29^2\), for TN and NT is \(2 * 0.29 * 0.71\) and for NN is \(0.71^2\).
03

Find probabilities for specific scenarios

To find the probability that neither pedestrian is texting (NN), use the probability found earlier - \((0.71)^2\). The probability that both pedestrians are texting (TT), would be \((0.29)^2\). And, the probability that exactly one pedestrian is texting (TN or NT), would be the sum of probabilities for TN and NT - \(2 * 0.29 * 0.71\).

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Key Concepts

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

Independence in Probability
When dealing with random events, understanding the concept of independence is crucial. Two events are considered to be independent if the occurrence of one does not affect the probability of the occurrence of the other. For instance, in the case of our pedestrians, whether one person texts while walking has no bearing on the likelihood of the other person doing the same.

To calculate the probability of two independent events both occurring, we multiply the probability of each event happening separately. This is illustrated in our textbook example by the calculation for both pedestrians texting: the probability of the first pedestrian texting is 29%, and since the events are independent, the probability of the second also texting is still 29%, leading to a combined probability of \(0.29 \times 0.29\).

Key Takeaway: Independence in probability allows us to easily calculate the likelihood of multiple events occurring together by simply multiplying their separate probabilities, provided that the events do not influence each other.
Probability of Events
The probability of an event is a measure of the likelihood that the event will occur. It is calculated by taking the number of ways the event can occur and dividing it by the total number of possible outcomes. Probabilities are expressed as numbers between 0 and 1, or between 0% and 100%. An event that cannot occur has a probability of 0, while an event that is certain to occur has a probability of 1 (or 100%).

In our pedestrian example, to find the probability that neither pedestrian texts while walking, which is the NN sequence, we calculated \(0.71^2\) because there is a 71% chance (100% - 29%) for each pedestrian not to text. The solution correctly utilized the fundamental rule that the probability of an event occurring is equal to one minus the probability of it not occurring.

Key Takeaway: Accurate determination of event probabilities forms the foundation of solving more complex probability problems, and always pay attention to whether events are independent or not, as it changes the approach to calculating their probabilities.
Combinatorial Probability
When the outcomes of an event can be combined in various ways, combinatorial probability becomes a handy tool to have. It involves counting the number of ways an event can occur, using combinations and permutations, where order matters for permutations but not for combinations.

In scenarios like our textbook problem where we look at two occurrences — a pedestrian texting or not — combinatorial probability helps us calculate scenarios such as 'exactly one of the pedestrians texting'. Since the order in which the pedestrians are selected does not affect the outcome, we can calculate the probability of one texting and one not texting as a combination of the two probabilities, which results in \(2 \times 0.29 \times 0.71\) because there are two ways this can happen (TN or NT).

Key Takeaway: Combinatorial principles apply to many probability problems and offer a systematic way to count the number of desirable outcomes, thereby allowing for the correct calculation of probabilities in more complex scenarios.

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