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Question #41072

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∫0x(t−4)3(t−5)3 dtf(x) = \int_0^x (t-4)^3 (t-5)^3 \,dt. Let aa be the point of local maximum and bb be the point of local minimum of f(x)f(x). A biased coin with probability of heads p=56p = \frac{5}{6} is tossed nn times. Let XX be the number of heads obtained. It is given that the probability mass function P(X=k)P(X=k) attains its maximum value at exactly two points, k=ak=a and k=bk=b. Find the conditional expectation E[X∣X<a]E[X \mid X < a].

Options

  • A.

    42751526\frac{4275}{1526}

  • B.

    2125763\frac{2125}{763}

  • C.

    31251526\frac{3125}{1526}

  • D.

    42757776\frac{4275}{7776}

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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

Complementary Counting|Random Variable Expectation|Second Derivative Test Limitations

Why This Question Is Tricky

Common Mistakes

This question tests the limitations of the Second Derivative Test and the conceptual understanding of probability mass functions.

Students often reflexively use f′′(x)f''(x) to classify extrema; here, f′′(4)=f′′(5)=0f''(4)=f''(5)=0, which acts as a trap, potentially leading them to misclassify the points as mere inflections.

Once a=4a=4 and b=5b=5 are correctly identified via the First Derivative Test, the problem shifts to a binomial distribution trap.

The PMF having a 'flat region' (two consecutive maxima) forces the deduction of the number of trials n=5n=5.

Finally, calculating the conditional expectation requires complementary counting to efficiently evaluate the restricted sample space, seamlessly weaving calculus, combinatorics, and probability.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptComplementary Counting
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree