Assume that you have 5 red balls and 4 blue balls from which randomly to select two without replacement. (1) What is the probability that exactly one red ball is chosen? (2) What is the probability that the second ball is blue given that at least one of the balls is blue?
there are a couple ways to do this perhaps the most intuitive way is to calculate the probability that the first ball is red and the second is not then the probability that the first is not red, and the second is do you know how to do that?
i just got the first part but the second part I'm not sure how to do
guess we need some notation
actually a diagram would be bets lets see if i can draw one
ok
actually no, lets just do it easy you have already computed the probability that exactly one ball is blue right?
what did you get, and how did you do it?
I got 5/9
the probability that both balls are blue is \[\frac{4}{9}\times \frac{3}{8}=\frac{1}{6}\]
the probability that the first is blue and the second is red is \[\frac{4}{9}\times \frac{5}{8}=\frac{5}{18}\]
and the probability that the first is red and the second is blue is the same \[\frac{5}{18}\] add these up to get the probability that at least one is blue
you need that for your denominator
so 1/6+5/18
no it is \[\frac{1}{6}+\frac{5}{18}+\frac{5}{18}\]
oh
the answer would be 13/18
ok good so that is your denominator
your numerator is the probability that the second ball is blue
so 4/9
yes
8/13?
that is what i get, yes
:D it was right thanks man.
yw nothing like instant gratification !
btw if i want to send you a message how do i do that?
you cannot
;(
you can tag me in a question if you like @satellite73
okay I'm going to be posting a lot of problems tonight because I have a midterm tomorrow that I have not even studied for yet and I pulled an all nighter last night to work on a presentation that I had to present today and study for a test that i took today.
yikes
@satellite73 I know. Right now I'm scared that I am going to have to stop doing my math hw and study for the test and take the fail on the hw
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