 Probability Maths Question??? Thanx to everyone who can help!? - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Probability Maths Question??? Thanx to everyone who can help!? 07-29-2007 #1 Ò³Ò²Ò³~Victory~Ò³Ò²Ò³ Guest   Posts: n/a Probability Maths Question??? Thanx to everyone who can help!? 1. Three balls are selected from a box containing 4 red balls and 5 blue balls. If the ball chosed after each selection is replaced before the next selection, find a) the probability distribution for the following number of red balls drawn:i. 0 red ballsii. 1 red balliii. 2 red ballsiv. 3 red balls(b) the probability that three reds are chosen, given that at least one ball is red.Please explain the steps as well...THANKS!!! because i have no clue on how to do it...and probability isn't my favorite topic! So thanx to everyone out there!  07-29-2007 #2 Phineas Bogg Guest   Posts: n/a (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. 07-29-2007 #3 TW K Guest   Posts: n/a Probability is one of my favs.a) I hope u know smoething about combinations: nCri_ 4C0 * 5C3 )/9C3Which means we are selecting none (0) balls out of four red balls and we are selecting 3 out of 5 blue balls.... The total probability will remain the same because we are slecting 3 random balls out of 9 given balls.ii_ 4C1*5C2)/9C3..iii_ 4C2*5C1)/9C3iv- _ 4C3*5C0)/9C3b: Now 3 balls are chosen and one of them is known to be red. So basically u now have to estimate the probability of selecting 2 red balls from 8 coloured balls because 1 ball is chosen and is known to be red. which will as in previous cases be of course 3C2*5C0)/ 8C2Hope it wz some helpTW K 07-29-2007 #4 qwert Guest   Posts: n/a X = number of red balls drawnP(X=0) = C(5,3) / C(9,3)P(X=1) =[ C(4,1) * C(5,2) ] / C(9,3)P(X=2) =[ C(4,2) * C(5,1) ] / C(9,3)P(X=3) = C(4,3) / C(9,3) rbmix.com

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