Question

The probability that two randomly selected subsets of the set $\left\{1,2,3,4,5\right\}$ have exactly two elements in their intersection, is:

• A. $\frac{65}{2^7}$
• B. $\frac{135}{2^9}$
• C. $\frac{65}{2^8}$
• D. four possibilities$\frac{35}{2^7}$

Answer 1

The correct option is B

$\frac{135}{2^9}$

Explanation for the correct option:

Step 1: Calculate the sample space

Given set is $\left\{1,2,3,4,5\right\}$

Let $P$ and $Q$ be the two subsets.

The total number of subsets for a set is $2^n$ where $n$ is the number of elements in the set. For the given set, $n=5$.
Thus, the total number of subsets for the given set is $2^5$

Since $2$ subsets are formed, the total number of sample space would be $2^5\times 2^5=2^{10}$

Step 2: Calculate the number of desired outcomes

For the intersection of the subsets to have two elements, the subsets themselves must have at least two elements.
For there to be an intersection with only two elements, the subsets must have only two common elements.

Number of ways to choose $2$ elements from $5$ elements is ${}^{5}C_2$.

Now, there remain $3$ elements whose fates aren't decided yet. The other $3$ elements can either be in $P$, or in $Q$, or in neither, but they can't be in both because then the intersection will have $3$ elements.
Thus, the total number of possibilities here is $3^3$.

Therefore, the total number of desired possibility is ${}^{5}C_2\times 3^3$

Step 3: Calculate the probabaility:

Probability is calculated as,

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

Where $P\left(A\right)$ is the probability of event $A$, $n\left(A\right)$ number of possibility of $A$ occuring, and $n\left(S\right)$ is the number of sample space

Thus, the required probability is,

$\begin{array}{rcl}P\left(A\right)& =& \frac{{}^{5}C_2\times 3^3}{2^{10}}\\ & =& \frac{5\times 2\times 27}{2^{10}}\\ & =& \frac{135}{2^9}\end{array}$

Hence, option B is correct.