Probability 5 Question 15
15. $A$ is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$. $A$ subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.
(1991, 4M)
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Answer:
Correct Answer: 15. $\frac{3}{4}{ }^{n}$
Solution:
- Since, set $A$ contains $n$ elements. So, it has $2^{n}$ subsets. $\therefore \quad$ Set $P$ can be chosen in $2^{n}$ ways, similarly set $Q$ can be chosen in $2^{n}$ ways.
$\therefore \quad P$ and $Q$ can be chosen in $\left(2^{n}\right)\left(2^{n}\right)=4^{n}$ ways.
Suppose, $P$ contains $r$ elements, where $r$ varies from 0 to $n$. Then, $P$ can be chosen in ${ }^{n} C _r$ ways, for 0 to be disjoint from $A$, it should be chosen from the set of all subsets of set consisting of remaining $(n-r)$ elements. This can be done in $2^{n-r}$ ways.
$\therefore \quad P$ and $Q$ can be chosen in ${ }^{n} C _r \cdot 2^{n-r}$ ways.
126 Probability
But, $r$ can vary from 0 to $n$.
$\therefore$ Total number of disjoint sets $P$ and $Q$
$$ =\sum _{r=0}^{n}{ }^{n} C _r 2^{n-r}=(1+2)^{n}=3^{n} $$
