Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (eujc21):

Suppose that 55% of all adults regularly consume coffee, 45% regularly consume carbonated soda, and 70% regularly consume at least one of these two products. a) What is the probability that a randomly selected adult regularly consumes both coffee and soda? b. What is the probability that a randomly selected adult doesn’t regularly consume at least one of these two products?

OpenStudy (eujc21):

a) P(A N B) = P(A) + P(B) - P(AUB) = .3

OpenStudy (eujc21):

for b. I'm stuck so far P(AUB)' but isnt that just 1-.7 ????

OpenStudy (dumbcow):

hey you got it....part b) is just 0.3 as well

OpenStudy (eujc21):

I'll post one more that I was stuck on for a while

OpenStudy (dumbcow):

Ven diagrams can help with these type of problems

OpenStudy (eujc21):

How would I go about this one \[P(A_1' & A_2' U A_3)\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!