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 October 12th, 2013, 02:06 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 A box contains 3 white, 8 red, 9 blue Hi 1) A box contains 3 white, 8 red, 9 blue balls. If 3 balls are selected at random, find the probability A) one ball of the each colour is selected B) all balls are of the same colour C) one red and two of any colour D) All 3 are red E) at least 1 is white F) the balls are drawn in the order red, white, blue. H) Two Balls are red 2) A box contain 4 bad 6 good tubes. Two are drawn from the box at a time. What is the probability A) that both the tubes drawn are good? B) one is good and other is bad?
 October 12th, 2013, 03:49 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: A box contains 3 white, 8 red, 9 blue I elaborate you the 2B, the others are similar. definition of the probability in general:p = number of the questioned cases / number of all cases in this case:p = number of one good and one bad drawn / number of 2 balls drawn from 10 that is: $p= \frac{4*6}{\binom{10}{2}}=\frac{24}{10*9/2}=24/45$
 October 12th, 2013, 05:12 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: A box contains 3 white, 8 red, 9 blue For the first question, let's begin with part A). How many ways are there to draw one ball of each color? How many ways are there in total to choose 3 balls from the total of 20?
 October 12th, 2013, 01:54 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,476 Thanks: 2039 2(A) In this earlier topic, the probability that both tubes drawn are good was found to be 1/3. 2(B) Similarly, the probability that both tubes drawn are bad is 2/15, so the required probability is 1 - 1/3 - 2/15 = 8/15.

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