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Question #41083

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=3x4−16x3+24x2f(x) = 3x^4 - 16x^3 + 24x^2 be a function defined on the sample space Ω=[−1,3]\Omega = [-1, 3]. A continuous random variable XX takes values in Ω\Omega with a probability density function p(x)=x+18p(x) = \frac{x+1}{8}. Let x1x_1 and x2x_2 be the points in Ω\Omega where f(x)f(x) attains its global minimum and global maximum, respectively. Let AA be the event that X∈[min⁡(x1,x2),max⁡(x1,x2)]X \in [\min(x_1, x_2), \max(x_1, x_2)], and BB be the event that XX falls in the interval where the graph of f(x)f(x) is strictly concave downward. Calculate the probability P(A′∩B′)P(A' \cap B').

Options

  • A.

    79144\frac{79}{144}

  • B.

    65144\frac{65}{144}

  • C.

    12\frac{1}{2}

  • D.

    116\frac{1}{16}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

De Morgan’s Laws in Probability|Global Extrema on Intervals|Second Derivative Test Limitations

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to synthesize global extrema analysis, concavity, and De Morgan's laws within a continuous probability framework.

The primary trap lies in the evaluation of the global maximum.

Students often rely exclusively on the Second Derivative Test to classify extrema.

For f(x)f(x), f′′(2)=0f''(2) = 0, making the test inconclusive.

A common failure mode is incorrectly assuming x=2x=2 is a local/global maximum without checking the boundaries of Ω=[−1,3]\Omega = [-1, 3].

If a student neglects the boundary x=−1x=-1 (where the true global maximum f(−1)=43f(-1)=43 occurs) and assumes the maximum is at x=2x=2, they incorrectly define Event AA as [0,2][0, 2].

This leads to P(A)=1/2P(A) = 1/2 and a cascading error resulting in the trap answer 1/21/2.

Furthermore, the disjoint nature of the correct intervals requires careful application of De Morgan's law P(A′∩B′)=1−P(A∪B)P(A' \cap B') = 1 - P(A \cup B).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptDe Morgan’s Laws in Probability
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree