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Question #41055

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the sample space of all possible outcomes when a fair coin is tossed $10$ times. Let $X(\omega)$ denote the number of heads in an outcome $\omega \in S$ . Define an event $E = \{ \omega \in S : (X(\omega)-3)(X(\omega)-6)^2(X(\omega)-8) \ge 0 \}$ . The probability $P(E)$ is equal to

Options

A.
$\frac{221}{512}$
B.
$\frac{29}{128}$
C.
$\frac{551}{1024}$
D.
$\frac{67}{1024}$

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

Algebra of Events|Properties of Binomial Coefficients|Even Multiplicity Roots

Why This Question Is Tricky

Common Mistakes

This question tests the integration of binomial probability sample spaces with the wavy curve method (method of intervals), specifically targeting the 'Even Multiplicity Root' trap.

Students often blindly alternate signs across all roots or divide out the non-negative term $(X-6)^2$ without considering the boundary case where it equals zero.

This rote application causes them to incorrectly exclude the isolated point $X=6$ from the solution set, leading directly to the trap option 29/128.

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