Question

Let A, B, C be subsets of the Universal set U. If $n\left( U \right)=692,n\left( B \right)=230,n\left( C \right)=370,n\left( B\cap C \right)=20,n\left( A\cap B'\cap C' \right)=10$, find $n\left( A'\cap B\cap C' \right)$.

Answer 1

For solving this question, we should know about the formulas of intersection and union and we should also know about the rules of probability for solving these types of questions related to probability. The union of two or more sets is the set that contains all the elements of the two or more sets. Union is denoted by the symbol $\cap$.

Complete step by step answer:
Here in the given question, the given terms are as follows, $n\left( U \right)=692,n\left( B \right)=230,n\left( C \right)=370,n\left( B\cap C \right)=20,n\left( A\cap B'\cap C' \right)=10$. And we have to find the value of $n\left( A'\cap B\cap C' \right)$. By the formulas of intersection and union, we know that,
$n\left( A'\cap B'\cap C' \right)=n\left( \left( B'\cap C' \right)nA' \right)$
And we can also write it as follows,
$\begin{aligned} & n\left( A'\cap B'\cap C' \right)=n\left( {{\left( B\cup C \right)}^{'}}\cap A' \right) \\\ & \because B'\cap C'={{\left( B\cup C \right)}^{'}} \\\ \end{aligned}$
So, here if we solve this formula, then we will get,
$n\left( A'\cap B'\cap C' \right)=n\left( {{\left( B\cup C \right)}^{'}} \right)-n\left( {{\left( B\cup C \right)}^{'}}\cap A \right)$
And we will solve this term as follows,
$\begin{aligned} & n\left( A'\cap B'\cap C' \right)=n\left( U \right)-n\left( B\cup C \right)-n\left( A\cap {{B}^{'}}\cap C' \right)\ldots \ldots \ldots \left( 1 \right) \\\ & \because n\left( {{\left( B\cup C \right)}^{'}} \right)=n\left( U \right)-n\left( B\cup C \right) \\\ & \because n\left( {{\left( B\cup C \right)}^{'}}\cap A \right)=n\left( A\cap {{B}^{'}}\cap C' \right) \\\ \end{aligned}$
So, for solving the values, we will get,
$\because n\left( B\cup C \right)=n\left( B \right)+n\left( C \right)-n\left( B\cap C \right)\ldots \ldots \ldots \left( 2 \right)$
Putting all the values in equation (1) and then again putting them in equation (2), we get as follows,
$n\left( B\cup C \right)=230+370-20$
Now, by equation (1), we will get,
$\begin{aligned} & n\left( A\cap B'\cap C' \right)=692-230-370+20-10 \\\ & \Rightarrow n\left( A\cap B'\cap C' \right)=102 \\\ \end{aligned}$

So, the value obtained of $n\left( A\cap B'\cap C' \right)$ is equal to 102.

Note: The intersection formulas and the union formulas are very much necessary for solving these types of questions and they will be all expected to be used in these questions. So, the union and intersection’s basics are very much useful for solving the questions and getting the exact answer.