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Chapter 6: Problem 77

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 probabilities are as follows: P(TT) = 0.08 (or 8%), P(NN) = 0.50 (or 50%), P(TN or NT) = 0.42 (or 42%).

Step by step solution

01

Identify all possible sequences for two pedestrians

The possible sequences for two pedestrians can be listed as TT, TN, NT, and NN; where T represents a pedestrian who is texting, and N represents one who is not.
02

Compute the probability for each sequence

Assuming independence, the probabilities for each sequence are: P(TT) = 0.29 * 0.29, P(TN) = 0.29 * 0.71, P(NT) = 0.71 * 0.29, and P(NN) = 0.71 * 0.71.
03

Calculate the probability that neither pedestrian texts while walking

The probability that neither pedestrian texts while walking is equivalent to the sequence, NN. Thus, P(NN) = 0.71 * 0.71.
04

Calculate the probability that both pedestrians text

The probability that both pedestrians text while walking is equivalent to the sequence, TT. Thus, P(TT) = 0.29 * 0.29.
05

Calculate the probability that exactly one pedestrian texts

The probability that exactly one pedestrian texts while walking is the sum of the probabilities for the sequences TN and NT. Thus, P(TN, NT) = P(TN) + P(NT).

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