+71 A stick is broken at two points, chosen at random. If the length of the stick is 6 units, then what is the probability that all three resulting pieces are shorter than 5 units?
What I've tried so far:
I've treated the stick as a number line, from 0 to 6. x and y are my variables for those two points on the line. I know that the probability of the first point being within the range of 1 to 5 so that the resulting two pieces are shorter than 5 is 5/6. I don't know how to get the second probability or how to combine them. I might be doing this completely wrong, could I get some help? Thanks, Starry.
+44 Here is the solution!
Let the length of the stick be L=6 units. Let X and Y be the two points where the stick is broken. We can assume that X and Y are chosen independently and uniformly at random from the interval [0,6].
Without loss of generality, assume X≤Y. Then the three pieces have lengths X, Y−X, and 6−Y. We want to find the probability that X<5, Y−X<5, and 6−Y<5.
The total possible region for (X,Y) is the triangle defined by 0≤X≤6, 0≤Y≤6, and X≤Y. The area of this region is 21⋅6⋅6=18.
We need to find the region where the following conditions hold:
\begin{enumerate}
\item X<5
\item Y−X<5
\item 6−Y<5⟹Y>1
\end{enumerate}
Also, we have 0≤X≤Y≤6.
Let's consider the region defined by these inequalities in the XY-plane.
\begin{enumerate}
\item X<5
\item Y
\item Y>1
\item X≤Y
\end{enumerate}
We can graph these inequalities. The region of interest is defined by the intersection of these inequalities.
\begin{itemize}
\item Y>1
\item X<5
\item Y
\item Y≥X
\end{itemize}
The vertices of the region are:
\begin{itemize}
\item Intersection of Y=1 and Y=X: (1,1)
\item Intersection of Y=1 and X=5: (5,1)
\item Intersection of X=5 and Y=X+5: (5,10) (but Y≤6)
\item Intersection of X=5 and Y=6: (5,6)
\item Intersection of Y=6 and Y=X+5: (1,6)
\item Intersection of Y=6 and Y=X: (6,6)
\item Intersection of Y=X+5 and Y=1: impossible
\item Intersection of Y=X and Y=X+5: impossible
\end{itemize}
So we have the vertices (1,1), (5,1), (5,6), and (1,6).
The area of this region is a rectangle with vertices (1,1),(5,1),(5,6),(1,6), whose area is (5−1)(6−1)=4⋅5=20.
However, we need to consider the restriction Y
The region is a rectangle with vertices (1,1),(5,1),(5,6),(1,6).
The area of this region is (5−1)(6−1)=4⋅5=20.
However, we need to consider the condition X≤Y≤6. The region defined by the inequalities is a rectangle with vertices (1,1),(5,1),(5,6),(1,6).
We must take the intersection of X≤Y and the rectangle.
The region is the rectangle with vertices (1,1),(5,1),(5,6),(1,6), and the condition X≤Y.
The area of the region is the area of the rectangle minus the area of the triangle where Y
The area of the region is the area of the rectangle, which is (5−1)(6−1)=4⋅5=20.
However, we need to consider the area of the region where X≤Y. This region is a rectangle with vertices (1,1),(5,1),(5,6),(1,6). The area is 4×5=20.
But we need to consider the case where X≤Y≤6.
We are looking for the region where 1 <6, X<5, Y
The area of the region where X≤Y≤6 and 0≤X≤6 is 21×6×6=18.
The area of the region where 1≤Y≤6, X≤Y, X≤5, and Y≤X+5 is
∫51(y−1)dy+∫65(5)dy=[y22−y]51+5(6−5)=(252−5)−(12−1)+5=7.5+0.5+5=13
The probability is 13/18.
Final Answer: The final answer is 13/18.
I hope this helps!
