My Math Forum Probability That They Are One Red, One White And One Blue Ba

 July 10th, 2012, 02:20 PM #1 Senior Member     Joined: Jan 2012 Posts: 721 Thanks: 7 Probability That They Are One Red, One White And One Blue Ba Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9 white and 4 blue identical balls. What is the probability that they are one red, one white and one blue balls? Please I need working and explaination.
 July 10th, 2012, 06:44 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Probability That They Are One Red, One White And One Blu The probability of getting a red ball is: $P$$\text{red}$$=\frac{5}{18}$ The probability of getting a white ball is: $P$$\text{white}$$=\frac{9}{18}=\frac{1}{2}$ The probability of getting a blue ball is: $P$$\text{blue}$$=\frac{4}{18}=\frac{2}{9}$ Hence, the probability of getting all 3 is: $P$$\text{red AND white AND blue}$$=\frac{5}{18}\cdot\frac{1}{2}\cdot\frac{2}{ 9}=\frac{5}{162}$ Now we must consider that there are 3! = 6 ways to order the 3 balls, hence the final probability is: $P(X)=6\cdot\frac{5}{162}=\frac{5}{27}$
 July 10th, 2012, 09:27 PM #3 Senior Member     Joined: Jan 2012 Posts: 721 Thanks: 7 Re: Probability That They Are One Red, One White And One Blu @MarkFL The question demands that I evaluate the probability that they are one red, one white and one blue balls in the three draws held. They did not give the order which the draws will follow, I therefore assume it could be in order. My working: Total number of balls = 18 P(r) = 5/18 P(w) = 9/18 = 1/2 P(b) = 4/18 = 2/9 The probability that they are one red, one white and one blue balls = P(1st-r,2nd-w,3rd-b) Or P(1st-w,2nd-b,3rd-r) Or P(1st-b, 2nd-r,3rd-w) (5/18*1/2*2/9) + (1/2*2/9*5/1 + (2/9*5/18*1/2) = 10 + 10 + 10/324 = 30/324 = 5/54 But the answer I got did not match with the answer they provided for the question. Is my working wrong or did I not follow the right principle? Lets compare our working: in your own, they are six orders but in mine they are three orders instead of six. Your own ------> 60/324 = 5/27 My own ------> 30/324 = 5/54 What made you believe that they are six others instead of three? Or how do I know the number of orders in probability like this?
 July 10th, 2012, 09:51 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Probability That They Are One Red, One White And One Blu We have 3 choices for the first draw, 2 for the second and 1 for the last, giving 3! = 3·2·1 = 6 six permutations: RWB RBW WRB WBR BRW BWR
 July 10th, 2012, 11:50 PM #5 Senior Member     Joined: Jan 2012 Posts: 721 Thanks: 7 Re: Probability That They Are One Red, One White And One Blu Remeber, the three balls are drawn one after the other with replacement.
 July 10th, 2012, 11:54 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Probability That They Are One Red, One White And One Blu Yes, that doesn't change the number of permutations of the possible ordering.
 July 11th, 2012, 12:21 AM #7 Senior Member     Joined: Jan 2012 Posts: 721 Thanks: 7 Re: Probability That They Are One Red, One White And One Blu If that be the case, that implies that even if two balls of colours say white and yellow are drawn one after the other, wether with or without replacement, in finding the of probability of getting one white and one yellow will be held in this order --------> (two choices for the first draw) + (one choice for the second draw being the last) draw. i.e 2:1, suming up the total draws to 3. Isn't it?
 July 11th, 2012, 12:25 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Probability That They Are One Red, One White And One Blu No, we don't add, we multiply 2·1 = 2: WY YW
 July 11th, 2012, 12:55 AM #9 Senior Member     Joined: Jan 2012 Posts: 721 Thanks: 7 Re: Probability That They Are One Red, One White And One Blu I think it is important at this juncture that I inform you that your system of working is quite different from the working and answers provided by the people who set the question. Here is the list of possible answers based on how you do the workings. A. 5/102 B. 5/136 C. 5/162 D. 5/204 E. 5/243. It is unforturnate to note that your working did not fall into any of these answers. Here is their own working: Total number of balls = 18 P(1R, 1W, 1B) = 5/18 * 1/2 * 2/9 = 5/162 = option C. What do you have say about it?
 July 11th, 2012, 01:09 AM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Probability That They Are One Red, One White And One Blu That would be the correct answer if a particular order is required for the 3 colors. For example, if it is required to draw a red ball first, then a white ball second, then lastly a blue ball, then the probability 5/162. This was not made clear in the original statement of the problem. No order was implied, at least by my interpretation of the problem.

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# a bag contains 7 yellow balls and 5 red balls. one ball is taken from the bag at random and not replaced. a second ball is taken from the bag . determine the probability of both the balls are different colors

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