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7 September, 07:19

According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend and 88.09% stayed home. what is the probability that a randomly selected american stayed home or traveled by car over the 2011 memorial day weekend? p (stayed home or travelled by car) =

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  1. 7 September, 10:31
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    p (traveled by car) = 9.6%

    p (stayed home) = 88.09%

    p (stayed home or travelled by car) = p (stayed home) + p (traveled by car) - p (stayed home and traveled by car)

    p (stayed home and traveled by car) = 0%

    p (stayed home or travelled by car) = 88.09%+9.6%-0%

    p (stayed home or travelled by car) = 97.69%
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