Question

If \[P(A\cup B)=0.8\] and \[P(A\cap B)=0.3\], then \[P(\overline{A})+P(\overline{B})=\]
A. \[0.3\]
B. \[0.5\]
C. \[0.7\]
D. \[0.9\]

Answer 1

Hint: In the above question, we are to find the value of the sum of probabilities of not the occurrence of events $A$ and $B$. By using set theory (complement of sets) and addition on probability, the required probability is calculated.



Formula Used:A probability is the ratio of favourable outcomes of an event to the total number of outcomes. So, the probability lies in between 0 and 1.
The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
According to set theory, complement of union/intersection of two sets is equal to the intersection/union of complemented sets respectively. I.e.,
$\begin{align}
  & {{A}^{'}}\cap {{B}^{'}}=(\overline{A\cup B})=1-(A\cap B) \\
 & {{A}^{'}}\cup {{B}^{'}}=(\overline{A\cap B})=1-(A\cup B) \\
\end{align}$
According to set theory,
Union of sets is described as:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
Intersection of sets is described as:
\[A\cap B=\{x/x\in A\text{ and }x\in B\}\]
Symmetric difference is described as:
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]



Complete step by step solution:Here, we are given the value of the probability of occurrence of both events $A$ and $B$ i.e.,\[P(A\cup B)=0.8\]
The value of the probability of occurrence of the common part of the two events $A$ and $B$, i.e.,
\[P(A\cap B)=0.3\]
By the addition theorem on probability,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
We can write,
\[\begin{align}
  & P(\overline{A}\cup \overline{B})=P(\overline{A})+P(\overline{B})-P(\overline{A}\cap \overline{B}) \\
 & \Rightarrow P(\overline{A})+P(\overline{B})=P(\overline{A}\cup \overline{B})-P(\overline{A}\cap \overline{B})...(1) \\
\end{align}\]
Then,
\[\begin{align}
  & P(\overline{A}\cup \overline{B})=1-P(A\cup B) \\
 & \text{ }=1-0.8 \\
 & \text{ }=0.2 \\
\end{align}\]
And
\[\begin{align}
  & P(\overline{A}\cap \overline{B})=1-P(A\cap B) \\
 & \text{ }=1-0.3 \\
 & \text{ }=0.7 \\
\end{align}\]
On substituting the values of the expressions in equation (1),
\[\begin{align}
  & P(\overline{A})+P(\overline{B})=0.2+0.7 \\
 & \text{ }=0.9 \\
\end{align}\]



Option ‘D’ is correct

Note: In this question, the addition theorem on probability is applied for finding the required probability. By substituting the appropriate values, the required probability is calculated. Here we may go wrong with the complimented probability.