
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
\[
A.{\text{ }}\dfrac{{{}^{2n}{C_n}}}{{{2^{2n}}}} \\
B.{\text{ }}\dfrac{1}{{{}^{2n}{C_n}}} \\
C.{\text{ }}\dfrac{{1 \cdot 3 \cdot 5 \cdot ........... \cdot \left( {2n - 1} \right)}}{{{2^n}n!}} \\
D.{\text{ }}{\left( {\dfrac{3}{4}} \right)^n} \\
\]
Answer
591.9k+ views
Hint: In order to find the probability of the given condition; first consider some variable for the given set according to the number of elements, find the total number of possible combinations of distribution of elements in the given 2 subsets. And finally find the number of possible subsets which satisfy the given condition. Find the ratio of these numbers of combinations to find the probability.
Complete step-by-step answer:
Let us consider the set as $X = \left\{ {1,2,3,.......,n} \right\}$
Let any of the term from the set $X$ be $i$
As we know that there are 2 subsets of $X$ and they are $A$ and $B$
Then the number $i \in X$ has four distinct possibilities for the set A and set B. They are:
$
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 \\
$
But we know that there are n different elements in the set $X$
And also we have 4 different possibilities.
As we know that number of possible combination of m different items each having t possible presentation is ${m^t}$
Using the same formula we shall proceed further.
Therefore, there are ${4^n}$ ways in which we can randomly choose two subsets $A$ and $B$ of $X$ .
Now, on the basis of the above understanding, we will find the number of ways in which the given conditions get satisfied.
$A \cup B = X$
In order to satisfy the given condition, we need to satisfy one out of the given condition for every element $i \in X$
We need to have.
$
1.i \in A,i \in B \\
2.i \in A,i \notin B \\
3.i \notin A,i \in B \\
$
As we have “n” number of elements 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 $X$ such that $A \cup B = X$ is equal to
$\dfrac{{{3^n}}}{{{4^n}}} = {\left( {\dfrac{3}{4}} \right)^n}$
Hence, the probability that $A \cup B = X$ is ${\left( {\dfrac{3}{4}} \right)^n}$
So, option D is the correct option.
Note: Probability is the ratio of number of favorable outcomes in the event over the total number of possible outcomes in an event. In order to find the probability in such types of cases we just find the number of favorable sets by the use of the property of permutation and combination rather than using the fundamental method of counting.
Complete step-by-step answer:
Let us consider the set as $X = \left\{ {1,2,3,.......,n} \right\}$
Let any of the term from the set $X$ be $i$
As we know that there are 2 subsets of $X$ and they are $A$ and $B$
Then the number $i \in X$ has four distinct possibilities for the set A and set B. They are:
$
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 \\
$
But we know that there are n different elements in the set $X$
And also we have 4 different possibilities.
As we know that number of possible combination of m different items each having t possible presentation is ${m^t}$
Using the same formula we shall proceed further.
Therefore, there are ${4^n}$ ways in which we can randomly choose two subsets $A$ and $B$ of $X$ .
Now, on the basis of the above understanding, we will find the number of ways in which the given conditions get satisfied.
$A \cup B = X$
In order to satisfy the given condition, we need to satisfy one out of the given condition for every element $i \in X$
We need to have.
$
1.i \in A,i \in B \\
2.i \in A,i \notin B \\
3.i \notin A,i \in B \\
$
As we have “n” number of elements 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 $X$ such that $A \cup B = X$ is equal to
$\dfrac{{{3^n}}}{{{4^n}}} = {\left( {\dfrac{3}{4}} \right)^n}$
Hence, the probability that $A \cup B = X$ is ${\left( {\dfrac{3}{4}} \right)^n}$
So, option D is the correct option.
Note: Probability is the ratio of number of favorable outcomes in the event over the total number of possible outcomes in an event. In order to find the probability in such types of cases we just find the number of favorable sets by the use of the property of permutation and combination rather than using the fundamental method of counting.
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