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Question #41056

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, \dots, 21\}$ . A subset $A$ of $S$ is chosen uniformly at random, and let $X = |A|$ denote the number of elements in $A$ . Define two events $E_1$ and $E_2$ as follows: $E_1 : \sum_{k=0}^{10} \binom{X}{k} = 2^X$ $E_2 : \sum_{k=11}^{21} \binom{21-X}{k-X} = 2^{21-X}$ (Assume $\binom{n}{r} = 0$ for $r < 0$ or $r > n$ ). If $p = P(E_1 \cup E_2)$ and $q = P(E_1 \cap E_2)$ , find the ordered pair $(p, q)$ .

Options

A.
$(1, 0)$
B.
$\left(\frac{3}{4}, \frac{1}{4}\right)$
C.
$\left(1, \frac{1}{4}\right)$
D.
$\left(\frac{3}{4}, 0\right)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Mutually Exclusive vs Independent|Binomial Coefficient Summation Properties|The Empty Set Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Empty Set Trap' by disguising mutually exclusive events as independent ones using complex binomial summation boundaries.

Students often calculate P(E_1) = 1/2 and P(E_2) = 1/2, and falsely assume independence, leading to P(E_1 \cap E_2) = 1/4.

However, rigorous evaluation of the summation indices reveals that E_1 requires X \le 10 and E_2 requires X \ge 11.

The intersection is the empty set, making the events mutually exclusive (q=0) rather than independent, forcing a re-evaluation of P(E_1 \cup E_2) to 1 instead of 3/4.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Mutually Exclusive vs Independent
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free