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A stick is broken at two points, chosen at random. If the length of the stick is 6 then what is the probability that all three resulting pieces are shorter than 5 units?

 Mar 5, 2023
 #1
avatar+195 
+1

We can solve this problem using geometric probability. 

Imagine the stick as a line segment of length 6 on a coordinate plane, where one endpoint is at the origin (0,0) and the other endpoint is at (6,0). We can choose the two points where the stick is broken by selecting two random points on the line segment. The location of the first point can be represented by a number between 0 and 6, and the location of the second point can be represented by a number between the location of the first point and 6.

If we plot these two points on the coordinate plane, we get a rectangle with area 18 (since the base is 6 and the height is 3). The total number of ways to choose two points on the stick is equal to the area of this rectangle.

Now, we need to find the region of the rectangle that corresponds to the case where all three resulting pieces are shorter than 5 units. 

Suppose the first point is located at a distance x from the origin. Then, the second point can be located anywhere between x and 6. The length of the first resulting piece is x, the length of the second resulting piece is y - x (where y is the location of the second point), and the length of the third resulting piece is 6 - y. 

For all three resulting pieces to be shorter than 5 units, we need:

x < 5
y - x < 5
6 - y < 5

Rearranging these inequalities, we get:

0 < x < 5
x < y < x + 5
1 < y < 4

The region of the rectangle that corresponds to this set of inequalities is a trapezoid with base 4 and height 3. The area of this trapezoid is (1/2)(4 + 1)(3) = 7.5.

Therefore, the probability that all three resulting pieces are shorter than 5 units is equal to the area of the trapezoid divided by the area of the rectangle:

P = 7.5/18 = 5/12

So the probability that all three resulting pieces are shorter than 5 units is 5/12.

 Mar 5, 2023
 #3
avatar+16 
+2

Thank you so much for giving this question a shot! I'm afraid that was not the correct answer, but thank you for trying.

Capitalist  Mar 5, 2023
 #2
avatar
+1

Let the length of the first piece be $x$, the length of the second piece be $y$, and the length of the third piece be $6-x-y$. The total length of the stick is 6, so $0\leq x,y \leq 6$. The condition that all three resulting pieces are shorter than 5 units can be expressed as:

x<5andy<5and6<5x<5andy<5and6−x−y<5

Simplifying this inequality, we get:

x+y>1

 

The probability that all three resulting pieces are shorter than 5 units is equal to the probability that $x+y>1$ when $x$ and $y$ are chosen randomly from the interval $[0,6]$.

To visualize this, we can plot the region of the $xy$ plane where $x+y>1$ and $0\leq x,y\leq 6$:

 

 

 

The shaded region above the line $x+y=1$ represents the possible values of $x$ and $y$ that satisfy the condition $x+y>1$. The area of this region can be calculated as follows:

12(6−1)2=12.2521​(6−1)2=12.25

 

This is the total area of the square with side length 5 (i.e., the area of the region where $0\leq x,y<5$) minus the area of the triangle with base 5 and height 1/2 (i.e., the area of the region where $x+y<1$).

The total area of the square with side length 6 (i.e., the region where $0\leq x,y\leq 6$) is $6^2=36$. Therefore, the probability that all three resulting pieces are shorter than 5 units is:

12.25/36 = 49/144.

Therefore, the probability that all three resulting pieces are shorter than 5 units is 49/144..

 Mar 5, 2023
 #4
avatar+195 
+1

Yes, your solution is correct. The area of the region where $x+y>1$ is indeed 12.25, and the probability that all three resulting pieces are shorter than 5 units is 49/144. Your visual representation of the region is also helpful in understanding the problem. Well done!

Justingavriel1233  Mar 5, 2023

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