a box contains 2 red, 3 white, and 4 blue balls. if the 3 balls are drawn at random w/o replacement, determine the probability that are blue and 1 is white.
Jello, rcs!
A box contains 2 red, 3 white, and 4 blue balls.
If the 3 balls are drawn at random w/o replacement,
determine the probability that 2 are blue and 1 is white.
Selecting 3 balls from 9, there are: .$\displaystyle {9\choose3} = 84$ possible outcomes.
To get 2 blue and 1 white, there are: .$\displaystyle {4\choose2}{3\choose1} = 18$ ways.
Therefore: .$\displaystyle P(\text{2 blue, 1 white}) \;=\;\frac{18}{84} \;=\;\frac{3}{14}$
I had given a detailed analysis of how Soroban got those formula, but then decided, "Why bother. RCS has made it clear he doesn't really care about how to do the problem, he just wants the answer."
HallsOfIvy thanks for calling me " he " .... couldn't you also understand why i posted them because i evidently need help... cant you get my point why im here? ... what is this site for? for something everybody knew already??? wow if that is the case then this site should never be named methhelpforum... you may call this site as ... knownmathforallforum.com or neverpostmathproblem.com or don'tbotherHallsofIvy.com ... you sounded too good to be true. you are only human dont pretend like you knew all things, and besides i dont need your answer... i better have problems left unanswered by helper rather than correct answer from you. ive been posted my comment addressed to you that i dont need your help... you are not the only helper here.
Do me good by word ... and ill return all them to you.
God Bless