A stick of length L is to be broken into three parts. What is the prob

Question

A stick of length L is to be broken into three parts. What is the probability that the resulting pieces form a triangle?

(A)  \frac{1}{2}

(B)  \frac{1}{3}

(C)  \frac{1}{4}

(D)  1

(E) None of these.

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ATIQ12 · Answer

E,because of we don't know the piece of what type..

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BlueDefender · Answer

I'm not really certain about this reasoning but approaching it from the standpoint of limits:

Obviously "a" cannot be more than 1/2 of L because b+c>a

The probability of of there being a triangle when "a" approaches limit L/2 is 100% as that value provides the greatest possible range for values between b and c that would suffice a>c-b from the condition a+b>c.

If "a" is just near but slightly less than the center point of L (L/2) then that segment may be superimposed on the center of the remaining segment visually displaying the range of possible positions for breaks that would satisfy the condition of a>b-c, with infinitesimal tails at each each end where the condition would not be satisfied...
This would cover nearly the whole segment as "a" is approaching the limit L/2 i.e. you can break the remaining segment at virtually any point and it would result in a triangle.

As "a" approaches limit 0 of L however, the probability of there being a triangle also linearly decreases as it decreases the range of possible values b and c may take such that a>b-c.

so if you were to plot the probability of success vs. failure on a vertical axis with "a" representing the horizontal axis as a portion of L, you would see the probability of success, starting from 0 at "a" approaching limit 0 of L, linearly increasing up to limit L/2 where it reaches a peak of 100% and cutting off abruptly to 0 thereafter for the remainder until a=L.

The graph would represent the continuous distribution probability function and the total/cumulative probability would be calculated as area under graph for success /(area under graph for success+area under graph for failure) yielding 1/4.

I have to note that I may have reached the correct answer with faulty reasoning because I'm considering the possibility of "a" being greater than L/2, whereas in reality, it may be obvious that if you break it to either left or right of center that it would be the larger of the segments you would break the second time and not the smaller...I don't really know if such a thing should be factored in

I feel like there is some
Break 1 probability 1/2 because of x
and
Break 2 probability 1/2 because of y
giving a probability of 1/4 that I'm not seeing

I feel like there is some Break 1 probability 1/2 because of x and Break 2 probability 1/2 because of y giving a probability of 1/4 that I'm not seeing

hadimadi · Answer

Hi,

let a,b,c be the lengths with a+b+c=L.
We need to show that these things hold:

(1) a+b<c
(2) b+c<a
(3) a+c<b

Imagine creating the first length, a, by cutting the piece in half and taking a side- that won't work because then whatever b,c are, we get a=b+c -> (0 < a < L/2) and (L/2 < b+c < L).

This means that for the first cut, anything that leaves more than 50% of the stick is fine for us.

We need to still show that b+a < c and a+c > b. From the first, we can derive b-c < a. Let b>=c without loss of generality:

(b-c) < a, but we defined 0 < a < 0.5L -> (b-c) < 0.5L

Now imagine cutting the length of L-a=b+c into half. For every unit increase in one of the points, the other point has to decrease by the same amount, because (a-1unit)+(b+1unit)=L-a. But this implies that at maximum, we can increase either of the points by less than a/2 (otherwise, b-c>0.5L). Dividing L-a into the maximum possible lengths for b and c is

c=[(b+c)/2]-(a/2), b=[(b+c)/2]+(a/2),

Every cutting point that lies within the range c and and b is therefore a valid cut off point. This means that a/(b+c) of the area of b+c is a valid cut off point. This means that a/L points are valid cut off points for our second cut. With this method, we found a possible solution for any length 0 < a < L/2.

This smells like an integration problem, but since we are on the GMAT Forum, we should plot the possible solutions and pick smart numbers. Let L=1:

a=0.0000...01 -> b,c anything in the range of (0,0.5)
a=0.2 -> b,c are anything in the range of (0.3, 0.5) (because L-0.2=0.8, and a/2=0.1 -> (0.8/2)+/-0.1, and we get the ranges)
a=0.4 -> b,c are anything in the range of (0.3, 0.4)
a=0.499999..99 -> b,c are both about 0.5

If we plot this , we get what is the red area on the attached image, which are all valid points we can get for b,c (note that our picks for a are not on this two dimensional graph - rather, the red area represents ALL possibilities for b,c which satisfy our conditions) after picking any 0 < a < L/2 (in our case 1/2). The big blue triangle is the area of all possible outcomes for a+b+c=L=1. We can clearly see that the red area can exactly fit into the three areas which are indicated with a green arrow. In total, the blue area is 4 times the red area, so our result is 1/4.

As always no guarantees that this is correct, any comments?

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Mugdho · Answer

Bunuel 
 KarishmaB 
 chetan2u

What is the correct solution to this problem? Help plz

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bumpbot · Answer

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A stick of length L is to be broken into three parts. What is the prob

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A stick of length L is to be broken into three parts. What is the prob

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