+19 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 6 units?
I've seen the other posts, and I can't seem to figure it out.. can anyone help me? :(
+1222 To find the probability that all three pieces of a stick, which is 6 units long, are shorter than 6 units after it is broken at two random points, we can use a geometric interpretation.
1. **Setting Up the Problem:**
- Let the stick start at 0 and end at 6 (it has a total length of \( L = 6 \) units).
- Denote the two break points as \( X_1 \) and \( X_2 \), where \( 0 < X_1 < X_2 < 6 \).
- The resulting pieces of the stick will have lengths:
- Piece 1: from 0 to \( X_1 \) (length \( X_1 \))
- Piece 2: from \( X_1 \) to \( X_2 \) (length \( X_2 - X_1 \))
- Piece 3: from \( X_2 \) to 6 (length \( 6 - X_2 \))
2. **Condition for the Pieces:**
- We need each piece to be less than 6 units in length:
- \( X_1 < 6 \)
- \( X_2 - X_1 < 6 \)
- \( 6 - X_2 < 6 \)
- The inequalities simplify to:
- \( X_1 < 6 \) (which will always be true since \( X_1 < X_2 < 6 \))
- \( X_2 < 6 + X_1 \) (or simply \( X_2 < 6 \) due to initial constraints)
- \( X_2 > 0 \) (will also hold since \( X_2 > X_1 > 0 \))
3. **Valid Conditions:**
- The only actual restriction is that \( X_2 < 6 \) (which is always satisfied because \( 0 < X_1 < X_2 < 6 \)).
- The lengths of the pieces will always be less than the length of the original stick (6 units).
4. **Geometric Approach:**
- We represent the breaking points \( (X_1, X_2) \) in the coordinate plane (in the region where \( 0 < X_1 < X_2 < 6 \)).
- The total area of the triangle formed by these restrictions in this coordinate system is represented by:
\[
\text{Area of valid region} = \frac{1}{2} \times 6 \times 6 = 18
\]
- However, the actual region that must be considered is bounded by the triangle formed by \( (0,0), (6,0), (6,6) \). The area of the square is \( 6 \times 6 = 36 \).
5. **Calculating the Probability:**
- The area of the successful outcomes where all pieces are shorter than 6 units is the same as the area where the inequalities hold. As we analyzed, if we consider the constraints, the condition is satisfied for all relative placements of the breaking points.
- Thus the probability \( P \) of obtaining such a division where each piece is shorter than 6 units is simply:
\[
P = \frac{\text{area of valid choices}}{\text{total area}} = \frac{18}{36} = \frac{1}{2}
\]
Thus, the probability that all three resulting pieces are shorter than 6 units is \(\frac{1}{2}\) or 50%.
