Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Some red, white, and blue candies were placed in a bowl. Some contain nuts, and some do not. Suppose one of the candies were chosen randomly from all the candies in the bowl. According to the table below, if the candy is blue, what is the probability that it does not contain any nuts?

OpenStudy (anonymous):

OpenStudy (anonymous):

@openstudyneedsme

OpenStudy (anonymous):

10%

OpenStudy (camerondoherty):

If you add up all of the things values it adds up to 100

OpenStudy (anonymous):

its 10 percent

OpenStudy (anonymous):

wait wrong

OpenStudy (camerondoherty):

that makes it realtivaly easy to find out what it is

OpenStudy (camerondoherty):

so you can convert to a fraction to be 20/100

OpenStudy (anonymous):

20% cause its over half of 30

OpenStudy (camerondoherty):

since the value is 20 blue without nutes

OpenStudy (camerondoherty):

so tht would end up being 20%

OpenStudy (anonymous):

so it be 20%

OpenStudy (anonymous):

yes ^.^

OpenStudy (camerondoherty):

Yep c:

OpenStudy (anonymous):

ohh yup 20

OpenStudy (anonymous):

cause half of 30 is 15 and its not 15 its 20 so whats 10 plus 50?

OpenStudy (anonymous):

dis is confusing i am levinggg

OpenStudy (anonymous):

*sigh* didn't get a medal and i gave the first answer :'( waisted my time bruh

OpenStudy (anonymous):

aint going to help @bunniesforever anymore dont tag me

OpenStudy (mathmate):

@bunniesforever Have you done conditional probabilities?

OpenStudy (anonymous):

no >.> @mathmate

OpenStudy (mathmate):

ok, in that case, we know that the candy is blue, other colours do not come into consideration.

OpenStudy (mathmate):

What do you get for probability of NOT having nuts?

OpenStudy (mathmate):

@camerondoherty Thank you for answering, but I would wait till @bunniesforever answers before I pronounce the results!

OpenStudy (camerondoherty):

Sorry >.<

OpenStudy (anonymous):

tbh think @mathmate deserves the medal

OpenStudy (mathmate):

@camerondoherty You are most welcome to work on it together, but since @bunniesforever is the "owner" of the question, I really appreciate hearing from him/her.

OpenStudy (mathmate):

Well, since @bunniesforever seems to be busy, yes, @camerondoherty, you've got the correct answer! Thank you for helping!

OpenStudy (camerondoherty):

Lol

OpenStudy (camerondoherty):

Sorry about tht tho

OpenStudy (mathmate):

@camerondoherty nothing to be sorry about, we're all studying together. Too bad @bunniesforever is not available.

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!