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Question #41053

Difficulty: 8/10 | 🟢 Free

The Question

Let $X$ be a random variable following the binomial distribution $B\left(100, \frac{1}{2}\right)$ . Evaluate the sum: $S = \sum_{k=51}^{100} (k^2 - 100k + 2499)^2 \binom{100}{k} \left(\frac{1}{2}\right)^{100}$

Options

A.
$\frac{3627}{4} - \frac{1}{2^{101}} \binom{100}{50}$
B.
$\frac{3627}{4} - \frac{1}{2^{100}} \binom{100}{50}$
C.
$\frac{3725}{4} - \frac{1}{2^{101}} \binom{100}{50}$
D.
$\frac{3725}{4} - \frac{1}{2^{100}} \binom{100}{50}$

Check Your Answer

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|Binomial Distribution Tail-Ends|Polynomial Function Parity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize polynomial symmetry within the context of binomial tail-end probabilities.

A brute-force approach of expanding the polynomial into falling factorials (e.g., $k(k-1)(k-2)(k-3)$ ) to absorb the binomial coefficient $\binom{100}{k}$ is standard for full sums, but becomes computationally intractable for a half-sum (tail-end) because the upper limit does not allow the series to collapse cleanly.

The trap is failing to shift the index to the mean ( $k = r + 50$ ).

Doing so reveals the polynomial is perfectly even ( $r^4 - 2r^2 + 1$ ).

This parity allows the student to relate the intractable tail-end sum to the easily computable full expectation $E[P(X)]$ using the symmetry of the binomial distribution $B(100, 1/2)$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Algebra of Events
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free