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Pulsified333
 one year ago
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?
Pulsified333
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there 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?

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0i just got the first part but the second part I'm not sure how to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0guess we need some notation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually a diagram would be bets lets see if i can draw one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually no, lets just do it easy you have already computed the probability that exactly one ball is blue right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what did you get, and how did you do it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the probability that both balls are blue is \[\frac{4}{9}\times \frac{3}{8}=\frac{1}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the probability that the first is blue and the second is red is \[\frac{4}{9}\times \frac{5}{8}=\frac{5}{18}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need that for your denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no it is \[\frac{1}{6}+\frac{5}{18}+\frac{5}{18}\]

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0the answer would be 13/18

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok good so that is your denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your numerator is the probability that the second ball is blue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is what i get, yes

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0:D it was right thanks man.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yw nothing like instant gratification !

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0btw if i want to send you a message how do i do that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can tag me in a question if you like @satellite73

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0okay 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.

Pulsified333
 one year ago
Best ResponseYou've already chosen the best response.0@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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