Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
Two subsets $A$ and $B$ of a set containing $n$ elements are chosen at random. The probability that $A \subseteq B$ is:
$\frac{2^n}{2}$
$\frac{2^n}{n!}$
$\left(\frac{2}{3}\right)^n$
$\left(\frac{3}{4}\right)^n$
Step-by-Step Solution
Key Concept: Use the binomial expansion $(1+2)^n = 3^n$ to count favorable cases where at least one element is selected.
The favorable outcomes are all subsets containing at least one element: $\binom{n}{1}2^0 + \binom{n}{2}2^1 + \cdots + \binom{n}{n}2^0 = 3^n$. Total outcomes are $2^n \times 2^n = 4^n$, so the required probability is $\left(\frac{3}{4}\right)^n$.
Correct Answer: 4