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Question #41016

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be a set containing 5 distinct elements. Let $A$ be the family of all subsets of $S$ having an even number of elements, and $B$ be the family of all subsets of $S$ having an odd number of elements. A random sample of 16 subsets is chosen without replacement from the power set of $S$ . Let $E$ be the event that the sample contains an even number of subsets from $A$ , and $F$ be the event that the sample contains an even number of subsets from $B$ . What is the probability of $E \cup F$ ?

Options

A.
$\frac{1}{2} + \frac{\binom{16}{8}}{2\binom{32}{16}}$
B.
$\frac{3}{4}$
C.
$\frac{1}{2} - \frac{\binom{16}{8}}{2\binom{32}{16}}$
D.
$\frac{1}{2}$

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

Mutually Exclusive vs Independent|Vandermonde's Identity|Binomial Coefficient Summation Properties

Why This Question Is Tricky

Common Mistakes

This question targets the lethal misconception of treating dependent combinatorial selections as independent Bernoulli trials.

Students often intuitively guess $P(E) \approx 1/2$ and $P(F) \approx 1/2$ , then incorrectly apply the union formula for independent events $P(E \cup F) = P(E) + P(F) - P(E)P(F) = 3/4$ .

The trap exploits the fixed-sum constraint ( $X+Y=16$ ), which makes the events perfectly dependent ( $E=F$ ).

Furthermore, it tests the advanced application of Vandermonde's identity and alternating binomial sums to exactly evaluate the constrained sample space, punishing students who rely on intuition rather than rigorous combinatorial summation.

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