MATH SOLVE

2 months ago

Q:
# 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)=

Accepted Solution

A:

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%

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%