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A stick of length 10 is broken in one place. Find the probability that the longer piece is at least twice as long as the shorter piece and no more than 5 longer than the shorter piece. I've tried systems of equations, graphing stuff, I can't get far on this one. The correct answer is 1/6, but I get 1/10. Any help is gladly appreciated. Thanks.

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    $\begingroup$ If you show how you get 1/10, someone might spot an error that you can then avoid on future problems. $\endgroup$ – David K Feb 25 '17 at 3:16
  • $\begingroup$ It was essentially guess and check, it wasn't an answer I came to with confidence $\endgroup$ – Nick Brown Feb 25 '17 at 3:20
  • $\begingroup$ Not sure how you would "guess and check" a question like this. (Guessing is easy, checking is the tricky part.) $\endgroup$ – David K Feb 25 '17 at 3:21
  • $\begingroup$ Yeah it was a desperate attempt at something, I would just come up with any number, say 6.5 and see if that fits the criteria for the longer side, and kept doing it and I got a range of 6.5-7.5 that it works for, so 1/10 of the stick fits the criteria, 1/10 $\endgroup$ – Nick Brown Feb 25 '17 at 3:31
  • $\begingroup$ Ah, I see. That actually would work (at least sort of) if you had honed in on the range a little more tightly. The upper bound 7.5 is the correct maximum, and 6.5 is close to the minimum (it's actually a little too low: 6.5 < 2(3.5)). But the other problem was you divided by 10, as if any value 0 to 10 was equally likely; but since you were measuring the length of the longer piece, only values in the range 5 to 10 are possible. $\endgroup$ – David K Feb 25 '17 at 3:39
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For the sake of developing intuition, I'll approach this with minimal use of equations. In general, I think a solution like Théophile's that sets up well-constructed equations will more reliably get results.

The condition "at least twice as long" means the break cannot occur in the middle $1/3$ portion of the stick. The longer piece must have a length at least $20/3$, that is, at least $2/3$ of the entire stick.

The condition "no more than $5$ longer" means the break must be somewhere in the middle $5$ units of length of the stick. Otherwise the break would leave more than $15/2$ units in the longer piece and less than $5/2$ units in the shorter piece, which is a difference of more than $5.$

So out of all the places on the stick where the break might occur (which I assume is equally likely along the entire length of the stick), breaks anywhere within the $1/4$ of the length of the stick on either end will fail to meet the conditions, and so will breaks in the middle $1/3.$ The proportion of the stick that remains, where the conditions will be satisfied, is $$ 1 - \frac14 - \frac14 - \frac13 = \frac16. $$

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The lengths of the longer piece and shorter piece will be $5+x$ and $5-x$, respectively, for some $0 < x < 5$. If the longer piece is at least twice as long as the shorter, but no more than five units longer, then we have

$$2(5-x) \le 5+x \le 5 + (5-x),$$

which leads to

$$\frac53 \le x \le \frac52.$$

The probability that $x$ satisfies this condition is

$$\frac{\frac52-\frac53}5 = \frac16.$$

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  • $\begingroup$ Did you mean to put the entire thing over 10 at the bottom, since the entire sick has length 10? Because the correct answer is 1/6 $\endgroup$ – Nick Brown Feb 25 '17 at 3:18
  • $\begingroup$ @NickBrown No, I meant 5! Consider that $x$ varies between 0 and 10, not 0 and 5. The answer to the question as you have phrased it is 1/3. It would be different if you asked, for example, what is the probability that the left piece is at least twice as long as the right piece, but no more than five times as long. $\endgroup$ – Théophile Feb 25 '17 at 3:21
  • $\begingroup$ @DavidK Oh, I see. Thanks for catching that! $\endgroup$ – Théophile Feb 25 '17 at 3:22
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    $\begingroup$ @NickBrown It seems I misread the question at first—mea culpa. I've updated the answer, but note that the denominator didn't change; make sure you understand why it's 5 and not 10. $\endgroup$ – Théophile Feb 25 '17 at 3:25
  • $\begingroup$ Okay I see what you're referring to now, thanks for the clarification $\endgroup$ – Nick Brown Feb 25 '17 at 3:34
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Interesting, I'm getting an answer of 1/3, can someone correct my mistake?

I'm taking a look at a range that can be cut. We know that at 1/6, it is the smallest we can cut, since anything else will produce a stick that is more than 5 times the other one We also know that it can't be more that 5/6, for the same reason The stick also has to be smaller than 1/3 and bigger than 2/3 because if it is in the middle, we would not have one that is at least twice the other.

Putting those together, we have: $\dfrac{1}{6}\leq x\leq\dfrac{1}{3}$ or $\dfrac{2}{3}\leq x\leq\dfrac{5}{6}$

$2\left(\dfrac{1}{3}- \dfrac{1}{6}\right) = \dfrac{2}{6} = \dfrac{1}{3}$

Assuming the probability of selecting any point to cut is the same, the probability to cut based on the specific conditions is $\dfrac{1}{3}$

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  • $\begingroup$ I misread it the same way initially -- the mistake is that the longer stick can't be more than 5 units longer than the shorter one (rather than 5 times as long). $\endgroup$ – Especially Lime Feb 26 '17 at 8:17
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Another solution:

Let $L$ be the length of the longer, and $l$ be the length of the shorter piece.

1) $ 2l \leq L \leq 5l$ ;

2) $L + l = 10$.

Eliminating $l$ in these inequalities yields:

3) $6 \frac{2}{3} \leq L \leq 8 \frac{1}{3}$.

Fix the stick horizontally, say, and put a ruler below (with length markings on it).

Now consider the sets of points:

$A$ =: { $x : 0 \leq x \lt 6 \frac{2}{3}$ }, $x$ representing the length of the stick measured from the left.

$P(A)$ stands for the probability of breaking a stick of length $ \lt 6 \frac {2}{3}$.

Ratios : $6 \frac {2}{3} : 3 \frac {1}{3} = 20 : 10$, yields

$P(A) = 2/3$.

$B$ := { $y : 0 \leq y \leq 8 \frac {1}{3}$ }, $y$ representing the length of the stick measured from the left.

Finding ratios: $8 \frac {1}{3} : 1 \frac { 2}{3} = 25 : 5 = 5 : 1$.

$P(B) = 5/6$.

The set $ B$ \ $A$ = { $z : 6 \frac {2}{3} \leq z \leq 8 \frac {1}{3}$} represents the desired range of lengths.

4) $B = A \cup$( $B$ \ $A$), union of two disjoint sets.

$P(B$ \ $A) = P(B) - P(A)$.

With $P(B) = 5/6, P(A) = 2/3$, we get

$P(B$ \ $A) = 5/6 - 2/3 = 1/6$.

$Question: $

Assuming the stick not fixed at one end, is there a factor of 2, since the same reasoning applies to" the other end" which would double the probability?

Comments are welcome.

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