Tricky Logic/Segment Problem?

Question

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.

Answer 1

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 $$ 1 - \frac14 - \frac14 - \frac13 = \frac16. $$

Answer 2

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

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

Answer 3

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.

Answer 4

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 $\dfrac{1}{3}$