Question

Let $A$, $B$, $C$ are the three events in a probability space. Suppose that $P\left( A \right)=0.5$, $P\left( B \right)=0.3$, $P\left( C \right)=0.2$, $P\left( A\cap B \right)=0.15$, $P\left( A\cap C \right)=0.1$ and $P\left( B\cap C \right)=0.06$, then the maximum possible value of $P\left( A'\cap B'\cap C' \right)$ is
A. $0.31$
B. $0.25$
C. $0$
D. $0.26$

Answer 1

Hint: In the problem we have the values of $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( A\cap B \right)$, $P\left( A\cap C \right)$, $P\left( B\cap C \right)$. From these values we can establish a relation between $P\left( A\cup B\cup C \right)$ and $P\left( A\cap B\cap C \right)$. From this relation we can calculate the value of $P\left( A\cap B\cap C \right)$ in terms of $P\left( A\cup B\cup C \right)$. After getting the value of $P\left( A\cap B\cap C \right)$ we can find the value of $P\left( A'\cap B'\cap C' \right)$ from the equation $P\left( A'\cap B'\cap C' \right)=1-P\left( A\cap B\cap C \right)$. From this relation we can find the minimum value of $P\left( A'\cap B'\cap C' \right)$.

Complete step-by-step answer:
Given that,
$A$, $B$, $C$ are the three events in a probability space and $P\left( A \right)=0.5$, $P\left( B \right)=0.3$, $P\left( C \right)=0.2$, $P\left( A\cap B \right)=0.15$, $P\left( A\cap C \right)=0.1$ and $P\left( B\cap C \right)=0.06$.
We know that $P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( A\cap C \right)-P\left( B\cap C \right)+P\left( A\cap B\cap C \right)$
Substituting the all the values we have, then we will get
$\begin{align}
  & P\left( A\cup B\cup C \right)=0.5+0.3+0.2-0.15-0.1-0.06+P\left( A\cap B\cap C \right) \\
 & \Rightarrow P\left( A\cup B\cup C \right)=0.69+P\left( A\cap B\cap C \right) \\
\end{align}$
Now the value of $P\left( A'\cap B'\cap C' \right)$ can be calculated by
$\begin{align}
  & P\left( A'\cap B'\cap C' \right)=1-P\left( A\cup B\cup C \right) \\
 & \Rightarrow P\left( A'\cap B'\cap C' \right)=1-\left[ 0.69+P\left( A\cap B\cap C \right) \right] \\
\end{align}$
Using the distribution law of multiplication in the above equation, then we will get
$\begin{align}
  & P\left( A'\cap B'\cap C' \right)=1-0.69-P\left( A\cap B\cap C \right) \\
 & \Rightarrow P\left( A'\cap B'\cap C' \right)=0.31-P\left( A\cap B\cap C \right) \\
\end{align}$
$\therefore $ The maximum possible value of $P\left( A'\cap B'\cap C' \right)$ is $0.31$ when $P\left( A\cap B\cap C \right)$ will become zero.

So, the correct answer is “Option A”.

Note: In probability problems we can simply save our time by checking the given options. We know that the maximum value of probability of any event should be less than. So, when there is any option that is having a number that is greater than one, then we will simply neglect it in our calculations.