Probability
Probability
Allen Star Batch
Grade 12

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: For each element in the n-element set, count independent choices: element must be in both A and B, in B only, or in neither (3 choices per element). This gives 3^n favorable outcomes where A ⊆ B. Total outcomes = (2^n)² = 4^n since A and B are chosen independently from 2^n subsets each.
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

Master Probability with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free