According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend a

Question

3 years ago

7

According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend a

nd 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)=

Mathematics

1 answer:

Ksivusya [100] · Answer 1 · 2022-04-16T23:13:22+03:00

<span>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%

</span>p(stayed home or travelled by car)=88.09%+9.6%-0%
p(stayed home or travelled by car)=97.69%