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C.

This problem tests your ability to translate a probability question into a simple geometric exercise. To start, the probability of X being in any specific location can be expressed as:

\(\dfrac{number of locations for X in a region}{ number of all possible locations of X}\)


Next, you can think of Line P as sitting on a number line. You can then relate the points on the line numerically, where lower values are to the left and greater values are to the right.

You are given Line P containing Point A and Point B where A < B.Additionally, Point X is randomly placed on the Line such that AX = 2• BX.

Quantity A is the probability of X being in Region 2 between A and B (A < X < B).

To solve this, set up the equation: AB = (AX) + (BX) Geometric relationship of the points

AB= 2 • BX+BX Substitute the given AB= 3 • BX Solve

This demonstrates there is only one solution in Region 2. Note that it is not important to know what the solution is; rather, you are looking for the NUMBER of possible solutions.

Quantity B is the probability of X being in Region 3 to the right of B (X > B). To solve, set up the equation:

AX = AB + BX Geometric relationship of the points

2 • BX = AB + BX Substitute the given

BX = AB Solve

This demonstrates there is also only one solution in Region 3. There are a total of two possible solutions. Using our starting ratio to find the probability of X being in Region 2:

\(\frac{number of locations for X in Region 2}{number of all possible locations of X} = \frac{1}{2 }= 50\)% Solve for Quantity A

Since you know that there is also only one location in Region 3, Quantity B will also be 50%. QuantityA is therefore equal to Quantity B so the correct answer choice is C.