September 17th, 2018, 06:35 PM  #1 
Newbie Joined: Sep 2018 From: Korea Posts: 4 Thanks: 0  Probability of pulling out a white ball?
I'm an English teacher who is fifteen years out of practice on my maths ability, lol. A student that I'm tutoring asked me about this problem because she apparently had not been able to visit with her maths teacher today, but it's beyond what I remember how to do, and her exam is tomorrow. Would you help me with this? "There are three white balls in a pocket, a red ball (X), and a blue ball (Y). When you take out one ball from this pocket, the odds of a white ball coming out are onefourth, and the probability of a blue ball coming out is twothirds. What is the value of yx?" Thanks in advance. 
September 17th, 2018, 07:02 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,317 Thanks: 1230 
what do X and Y stand for?

September 17th, 2018, 07:07 PM  #3 
Newbie Joined: Sep 2018 From: Korea Posts: 4 Thanks: 0  The question doesn't say beyond X representing one red ball and Y representing one blue ball. In my mind that means there are five balls in the pocket altogether, but the chances of pulling out a white ball is supposedly 1/4 while it's 2/3 for a blue ball, even though there are supposed to be three white balls. So it doesn't make sense to me, but I'm guessing that this question is supposed to be testing if you understand the theory behind it? 
September 17th, 2018, 07:47 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549  Quote:
 
September 17th, 2018, 08:02 PM  #5  
Newbie Joined: Sep 2018 From: Korea Posts: 4 Thanks: 0  Quote:
So let's go with what you say with x red balls and y blue balls instead of x = 1 red ball and y = 1 blue ball, lol. And I assumed probability, but the translation makes it difficult for me to know for sure if it's odds or probability. I'm sorry, it's a bit of a bigger mess than I anticipated.  
September 17th, 2018, 08:04 PM  #6  
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  Quote:
Quote:
But that's not right either, it must be 1/5. Your student misremembered the question. Either that or her math teacher's insane.  
September 17th, 2018, 11:23 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,317 Thanks: 1230 
if we go with Jeff's assumption that X is the # of red balls and Y is the # of blue balls we have $\dfrac{3}{3+X+Y} = \dfrac 1 4$ and $\dfrac{Y}{3+X+Y} = \dfrac 2 3$ we can solve this to obtain $X=1,~Y=8$ and $YX=7$ 
September 18th, 2018, 03:27 PM  #8 
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7 
romsek is correct, but to put it even more simply: If there are 3 white balls and a 1 in 4 of chance of drawing one, then there has to be 4 x 3 or 12 balls in total. Then if there is a two thirds chance of drawing a blue ball, the number of blue balls has to be two thirds of 12, or 8. 8 +3 = 11, so there is only one other ball, which is the red ball. 8 blue balls; 1 red ball; 8 1 = 7. Sorry, but decades ago I spent a lot of time informally tutoring other kids in math during study hall. I got very accustomed to explaining 'math in plain English' (or as plain as possible). 
September 18th, 2018, 05:05 PM  #9  
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549  Quote:
 
September 18th, 2018, 05:37 PM  #10  
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7  Quote:
 

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ball, probability, pulling, white 
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