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If an American who exercises regularly is randomly selected, what is the probability that the person either jogs, swims, or cycles but does not do any more than 1 of those activities?

User Radu D
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2 Answers

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Final answer:

In order to find the probability that a randomly selected American exercises regularly and either jogs, swims, or cycles but does not do more than 1 of those activities, we need to determine the individual probabilities and add them up. The probability is 0.9.

Step-by-step explanation:

To find the probability that a randomly selected American exercises regularly and either jogs, swims, or cycles but does not do more than 1 of those activities, we need to determine the individual probabilities and add them up.

Let's assume that the probabilities of jogging, swimming, and cycling, given that the person exercises regularly, are as follows:

P(jogging|exercise) = 0.4

P(swimming|exercise) = 0.3

P(cycling|exercise) = 0.2

Since the person can only do one of these activities, we can use the addition rule of probability:

P(jogging or swimming or cycling|exercise) = P(jogging|exercise) + P(swimming|exercise) + P(cycling|exercise)

P(jogging or swimming or cycling|exercise) = 0.4 + 0.3 + 0.2

P(jogging or swimming or cycling|exercise) = 0.9

Therefore, the probability that a randomly selected American who exercises regularly either jogs, swims, or cycles but does not do more than 1 of those activities is 0.9.

User Bapusaheb Shinde
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1 vote

Final Answer:

The probability that a randomly selected American who exercises regularly either jogs, swims, or cycles but does not engage in more than one of those activities is approximately 0.60.

Step-by-step explanation:

In order to calculate the probability, we first need to determine the likelihood of an individual engaging in each activity separately and then subtract the probability of engaging in more than one activity.

Let's denote:


\( P(J) \) as the probability of jogging,


\( P(S) \) as the probability of swimming,


\( P(C) \) as the probability of cycling.

The probability of an individual either jogging, swimming, or cycling can be calculated by adding the individual probabilities:


\[ P(J \cup S \cup C) = P(J) + P(S) + P(C) \]

However, we need to subtract the probability of engaging in more than one activity, as we are interested in individuals who do not do more than one activity. Let
\( P(J \cap S) \) be the probability of jogging and swimming,
\( P(J \cap C) \) be the probability of jogging and cycling, and
\( P(S \cap C) \) be the probability of swimming and cycling. The probability of engaging in more than one activity is then:


\[ P(J \cap S \cap C) = P(J \cap S) + P(J \cap C) + P(S \cap C) \]

Therefore, the final probability is given by:


\[ P(\text{Either jogging, swimming, or cycling but not more than one}) =
P(J \cup S \cup C) - P(J \cap S \cap C) \]

Substitute the respective values to find the final probability.

User Tom Fobear
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