Question

Travel agents collected data from recent travelers about their modes of transportation for their vacations, They found that 37% traveled by airplane, 8% traveled by train, and 7% traveled by airplane and train. Let A be the event that the mode of travel was airplane and let T be the event that the mode of travel was train.

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Answer 1

37% traveled by airplane

This means the probability of traveling by airplane (P(A)) is 0.37

8% traveled by train

This means the probability of traveling by train (P(T)) is 0.08

7% traveled by airplane and train

This double counts the people who traveled by both airplane and train.

We need to subtract this 7% from both the airplane and train percentages to get the correct probabilities.

So the corrected probabilities are:

P(A) = 0.37 - 0.07 = 0.30

P(T) = 0.08 - 0.07 = 0.01

Let's verify that these corrected probabilities add up to 1 (100%):

P(A) + P(T) = 0.30 + 0.01 = 0.31

Since the problem states only 37% traveled by airplane and 8% by train, with 7% by both, the remaining 48% must have traveled by other means.

So we can add that to get a total probability of 1:

P(A) + P(T) + P(other) = 0.30 + 0.01 + 0.48 = 0.79

Therefore, the corrected probabilities are:

P(A) = 0.30

P(T) = 0.01

P(other) = 0.48