Question

Let $X$ be a set containing n elements. Two subsets $A$ and $B$ of $X$ are chosen at random, the probability that $A\cup B=X$ is

• A. $\frac{{}^{2n}{C}_n}{2^{2n}}$
• B. $\frac{1}{{}^{2n}{C}_n}$
• C. $\frac{1\cdot 3\cdot 5\cdot ....(2n-1)}{2^{n}n!}$
• D. ${\left(\frac{3}{4}\right)}^n$

Answer 1

The correct option is D ${\left(\frac{3}{4}\right)}^n$

Let $X=\{1,2,\dots ,n\}$
Any number $i\in X$ has four distinct possibilities:
1. $i\in A,i\in B$
2. $i\in A,i\notin B$
3. $i\notin A,i\in B$
4. $i\notin A,i\notin B$
There are $n$ numbers in set $X$.

Therefore, there are $4^n$ ways in which we can randomly choose two subsets $A$ and $B$ of $X$.
Now we will count the number of ways in which $A\cup B=X$

In this case, we want every number $i\in X$ which satisfy one of the three

1. $i\in A,i\in B$
2. $i\in A,i\notin B$
3. $i\notin A,i\in B$

There are $n$ numbers in set $X$.

Therefore, there are $3^n$ ways in which we can randomly choose two subsets $A$ and $B$ of $X$ such that $A\cup B=X$.
So, the probability that two randomly chosen subsets $A$ and $B$ of $X$ satisfy $A\cup B=X$ is equal to $\frac{3^n}{4^n}={\left(\frac{3}{4}\right)}^n$.