Math Brainteaser: Probability of Longer Piece Length

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The discussion revolves around calculating the probability that the longer piece of a string, cut at a random point, is at least x times the length of the shorter piece. Participants emphasize the distinction between continuous and discrete probability distributions, noting that this problem involves infinite cutting points rather than a finite set. They suggest starting with simpler cases to understand the concept better, particularly focusing on the ratio of favorable outcomes to total outcomes. The probability is determined by the length of segments where the condition holds compared to the total length of the string. Understanding these principles is crucial for solving the problem effectively.
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Homework Statement


A piece of string is cut into two pieces at a randomly selected point. What is the probability that the longer piece is at least x times as long as the shorter piece?

Homework Equations


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The Attempt at a Solution


everyone in my family tried, but we couldn't figure it out.
this problem is strange, because usually probability deals with picking something out of a sum total.

thanks for any help :biggrin:
 
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Your family trying doesn't count so much as you trying. What did you try? Try a simple case first. What's the probability you cut the string in such a way that one piece is at least twice the length of the other? What's the measure of places where you can cut the string so it works versus the total measure of places on the string?
 
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oceanflavored said:
usually probability deals with picking something out of a sum total.

That is the case when we deal with Discrete Probability distribution. In this case, as you can see, there are infinite points to chose from, there isn't a set of finite points that are favorable. Rather, there is a 'range' of lengths that is favorable. As such, you are dealing with a Continuous Probability distribution.

and as Dick said already, we can't help you further unless you show us something.
 
Continuous versus discrete really aren't that different. And rohanprabhu is absolutely correct. The probability is the ratio of the length of the segments of points that work over the length of the whole segment. I sort of regret using the word 'measure' instead of 'length'. But it's really the same thing as taking the ratio of counts.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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