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September 17th, 2018, 06:35 PM   #1
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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 one-fourth, and the probability of a blue ball coming out is two-thirds. What is the value of y-x?"

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September 17th, 2018, 07:02 PM   #2
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what do X and Y stand for?
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September 17th, 2018, 07:07 PM   #3
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what do X and Y stand for?
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?
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September 17th, 2018, 07:47 PM   #4
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Quote:
Originally Posted by Ratatat View Post
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?
The question as posed does not make any sense to me either. But perhaps what is meant is that there are x red balls and y blue balls. And are you talking about probabilities or odds?
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September 17th, 2018, 08:02 PM   #5
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The question as posed does not make any sense to me either. But perhaps what is meant is that there are x red balls and y blue balls. And are you talking about probabilities or odds?
That's probably true. The question was originally written in Korean, and a translation app was used to put it in English.

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.
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September 17th, 2018, 08:04 PM   #6
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Quote:
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"There are three white balls in a pocket, a red ball (X), and a blue ball (Y).
Ok, there are five balls in the pocket.

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Originally Posted by Ratatat View Post
When you take out one ball from this pocket, the odds of a white ball coming out are one-fourth,
No they're not, they're 3/5. There are five balls, three of which are white.

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Originally Posted by Ratatat View Post
and the probability of a blue ball coming out is two-thirds.
But that's not right either, it must be 1/5.


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What is the value of y-x?"
Your student misremembered the question. Either that or her math teacher's insane.
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September 17th, 2018, 11:23 PM   #7
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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 $Y-X=7$
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September 18th, 2018, 03:27 PM   #8
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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).
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September 18th, 2018, 05:05 PM   #9
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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).
Explaining math in plain English is great for starting out. But natural languages have far more ambiguity than the artificial language of math, and the formalism of mathematics lets you tackle problems far too complex to render into comprehensible English. Plain English lets you see why the formalism is correct for the simple problems and thus gives the student confidence to adopt that formalism.
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September 18th, 2018, 05:37 PM   #10
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Quote:
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Explaining math in plain English is great for starting out. But natural languages have far more ambiguity than the artificial language of math, and the formalism of mathematics lets you tackle problems far too complex to render into comprehensible English. Plain English lets you see why the formalism is correct for the simple problems and thus gives the student confidence to adopt that formalism.
Sorry. I'll avoid chiming in in the future.
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