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Question #41052

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the sample space of all positive integer solutions (x,y,z)(x, y, z) to the equation x+y+z=50x + y + z = 50. A solution (x,y,z)(x, y, z) is chosen at random from SS. Let E1E_1 be the event that x,y,zx, y, z can form the side lengths of a triangle (including degenerate triangles where the vertices are collinear). Let E2E_2 be the event that x,y,zx, y, z can form the side lengths of a non-degenerate triangle (with positive area). Find the conditional probability P(E2∣E1)P(E_2 \mid E_1).

Options

  • A.

    2329\frac{23}{29}

  • B.

    2129\frac{21}{29}

  • C.

    2529\frac{25}{29}

  • D.

    2398\frac{23}{98}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Algebra of Events|Distributing Identical Items (Stars and Bars)|Triangle Inequality Mind Games

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of combinatorics (Stars and Bars) and probability (Algebra of Events) under geometric constraints.

The primary trap lies in the subtle distinction between degenerate (x+y≥zx+y \ge z) and non-degenerate (x+y>zx+y > z) triangles, which translates to a shift of just 1 in the boundary conditions (z≤25z \le 25 vs z≤24z \le 24).

Students often fail to recognize that the complementary events (e.g., z≥25z \ge 25) are mutually exclusive because the sum is constrained to 50, allowing for direct set subtraction without complex inclusion-exclusion.

Another common trap is calculating the absolute probability P(E2)P(E_2) instead of the conditional probability P(E2∣E1)P(E_2 \mid E_1).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptAlgebra of Events
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree