Question

The probability that at least one of the events \[A\] and \[B\] occurs is \[\dfrac{3}{5}\]. If and occur simultaneously with probability \[\dfrac{2}{5}\], then \[P(\overline{A})+P(\overline{B})\] is
A. \[\dfrac{2}{5}\]
B. \[\dfrac{4}{5}\]
C. \[\dfrac{6}{5}\]
D. \[\dfrac{7}{5}\]

Answer 1

Hint: In the above question, we are to find the value of the sum of \[P(\overline{A})\] and \[P(\overline{B})\]. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.



Formula Used:A probability is the ratio of favorable outcomes of an event to the total number of outcomes. So, the probability lies 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,
The union of sets is described as:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
The intersection of sets is described as:
\[A\cap B=\{x/x\in A\text{ and }x\in B\}\]
The 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 probability of the occurrence of at least one of the events \[A\] and \[B\], i.e., \[P(A\cup B)\]which is equal to \[\dfrac{3}{5}\], and the probability of the occurrence of events \[A\] and \[B\] simultaneously, i.e., \[P(A\cap B)\] which is equal to \[\dfrac{1}{5}\].

According to the addition theorem on probability of the events \[A\] and \[B\],
\[\begin{align}
  & P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
 & P(A\cup B)=[1-P(\overline{A})]+[1-P(\overline{B})]-P(A\cap B) \\
 & P(A\cup B)=2-[P(\overline{A})+P(\overline{B})]-P(A\cap B) \\
\end{align}\]
Thus,
\[\begin{align}
  & [P(\overline{A})+P(\overline{B})]=2-P(A\cup B)-P(A\cap B) \\
 & \text{ }=2-\dfrac{3}{5}-\dfrac{1}{5} \\
 & \text{ }=\dfrac{10-3-1}{5} \\
 & \text{ }=\dfrac{6}{5} \\
\end{align}\]



Option ‘C’ 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.