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anonymous
 5 years ago
An urn contains 3 red marbles and 2 blue marbles. Three marbles are drawn together. Let X be the number of red marbles drawn. a) List the probability function of X. b) Find the expected value of X, E(X).
anonymous
 5 years ago
An urn contains 3 red marbles and 2 blue marbles. Three marbles are drawn together. Let X be the number of red marbles drawn. a) List the probability function of X. b) Find the expected value of X, E(X).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an urn is what they out the ashes of dead ppl in after they creamate them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0X = number of red marbles P(X) is defined on a domain of 1, 2, and 3 Note: 0 isn't in the domain because it isn't a possible outcome when drawing 3 marbles since there are only 2 nonred marbles. The definition of the P(X) is just the assignment of the probabilities to the 1, 2, and 3 \[P(1 red)=\frac{C(3,1)*C(2,2)}{C(5,3)}\] \[P(2 red)=\frac{C(3,2)*C(2,1)}{C(5,3)}\] \[P(2 red)=\frac{C(3,3)*C(2,0)}{C(5,3)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oops, the last one should be "P(3 red)"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you SO much. I had a basic idea of this. The rest would be to just plug in the probabilities in the E(X) formula right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0E(X) = 1*P(1)+2*P(2)+3*P(3) :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks, do you have an email or instant messaging program?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying to prepare for finals and there is a lot of probability stuff on this level that I do not understand.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would love to help but I don't have a lot of time. :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah... ok. Thanks anyway.
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