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Old 07-29-2007   #2
Phineas Bogg
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(a) the probability distribution for the following number of red balls drawn:i. 0 red balls(5/9)^3 = 125/729 since there is a 5/9 chance that each ball is blueii. 1 red ballYou have three choices for which ball is red, then for each choice the probability is: (4/9)*(5/9)^2, so the total is 3*(4/9)*(5/9)^2 = 300/729iii. 2 red ballsYou have three choices for which ball is blue, then for each choice the probability is: (4/9)^2*(5/9), so the total is 3*(4/9)^2*(5/9) = 240/729iv. 3 red balls(4/9)^3 = 64/729 since there is a 4/9 chance that each ball is red(b) the probability that three reds are chosen, given that at least one ball is red.= Prob(three reds chosen and at least one red chosen) / Prob(at least one red is chosen) = Prob(three reds chosen)/Prob(at least one red chosen) = 64/(300+240+64) = 16/(75 + 60 + 16)= 16/151Edit: Doc D is usually right on these math questions, but I think he has made a minor error in part b..., nevermind, he just fixed it.
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