Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (hba):

Help me with part b.

OpenStudy (hba):

OpenStudy (hba):

tree diag?

OpenStudy (callisto):

Maybe you can't have 1 black and 1 red too?!

OpenStudy (hba):

yes @Callisto

OpenStudy (callisto):

Hmm You are going to pick two balls only..

OpenStudy (hba):

Yes

OpenStudy (callisto):

P(X+Y<2) = P(1 black 1 green) + P(1 black 1 red) + P(2 green) I guess :|

OpenStudy (hba):

There can't be 1 black and 1 red as 1+1=2 It would be P(1 black 1 green)+P(1red 1 green)+P(all green)

OpenStudy (hba):

I just need a confirmation that it would be like this :/

OpenStudy (callisto):

Ok..

OpenStudy (hba):

so am i right?

OpenStudy (callisto):

I guess yes.. *PS: I'm terrible in probability!

OpenStudy (hba):

okay what would be the probability like. I mean is P(bg)=(3/8)*(3/8)

OpenStudy (callisto):

No?!

OpenStudy (hba):

then?

OpenStudy (callisto):

P(1 Black 1 green) = P(black then green) + P(green then black) P (black then green) = 3/8 * 3/7

OpenStudy (hba):

how do you know its with or without replacement?

OpenStudy (callisto):

I supposed it is without replacement.

OpenStudy (hba):

okay thanks.

OpenStudy (hba):

@Callisto What about part (e)?

OpenStudy (hba):

@ash2326

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!