Probability: There are three white balls in a pocket, a red ball (X), and a blue ball

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chingutee

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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?"

Thanks in advance.
 
chingutee said:
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?"
Click to expand...

I'm sorry for the long delay in replying; newcomers' questions often hide among others due to the delay in posting for moderation.

I don't understand the question. What do X and Y (and x and y) stand for? And did you really mean to use the word "odds", which is different from probability?

Perhaps you (or the author) meant to say this:

"There are three white balls in a pocket, x red balls, and y blue balls. When you take out one ball from this pocket, the probability of a white ball coming out is one-fourth, and the probability of a blue ball coming out is two-thirds. What is the value of y-x?"
​

If so, then the probability of a white ball is 3/(3+x+y) = 1/4, and the probability of a blue ball is y/(3+x+y) = 2/3. These two equations can be solved for x and y.
 
chingutee said:
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?"

Thanks in advance.
Click to expand...
Surely you have miswritten this. What you have written "a red ball (X), and a blue ball (Y)" says that there is exactly one red ball, labeled "X", and exactly one blue ball, labeled "Y". If that were actually the case, the probabilities given would be wrong- and it any case, it would make no sense to subtract "labels" (x- y). I thought that the problem really said "there are three white balls in a pocket, X red balls, and Y blue balls". However, in that case we get a fractional number of red and blue balls: one and a half red balls and seven and a half blue balls.

Please go back an reread the problem!
 
HallsofIvy said:
I thought that the problem really said "there are three white balls in a pocket, X red balls, and Y blue balls". However, in that case we get a fractional number of red and blue balls: one and a half red balls and seven and a half blue balls.
Click to expand...

I got a reasonable answer when I solved it under this interpretation.
 
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