Question

If $A$ and \[B\] are two events, then \[P(A\cup B)=\]
A. \[P(A)+P(B)\]
B. \[P(A)+P(B)+P(A\cap B)\]
C. \[P(A)+P(B)-P(A\cap B)\]
D. \[P(A)\cdot P(B)\]

Answer 1

Hint: In this question, we are to find the probability of the union of two events. This is nothing but proof of the addition theorem on probability. By using set theory, the required probability is calculated.


Formula Used:The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
According to set theory,
the 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
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]



Complete step by step solution:Consider two events $A$ and \[B\].
These events are shown by the Venn diagram as below:



From the diagram, we have
\[\begin{align}
  & A\cup B=S \\
 & (B-A)\cup (A\cap B)=B \\
 & (B-A)A\cap (B-A)=\Phi \\
 & A\cup (B-A)=A\cup B \\
 & A\cap (A\cap B)=\Phi \\
\end{align}\]
Then, from these we can apply probability on \[(B-A)\cup (A\cap B)=B\] and \[A\cup (B-A)=A\cup B\].
Applying probability to \[(B-A)\cup (A\cap B)=B\]:
\[\begin{align}
  & (B-A)\cup (A\cap B)=B \\
 & P(B)=P\left[ (B-A)\cup (A\cap B) \right] \\
 & \text{ }=P(B-A)+P(A\cap B) \\
 & \Rightarrow P(B-A)=P(B)-P(A\cap B)\text{ }...(1) \\
\end{align}\]
Applying probability to \[A\cup (B-A)=A\cup B\]:
\[\begin{align}
  & A\cup (B-A)=A\cup B \\
 & P(A\cup B)=P\left[ A\cup (B-A) \right] \\
 & \text{ }=P(A)+P(B-A) \\
\end{align}\]
On substituting (1), we get
\[P(A\cup B)==P(A)+P(B)-P(A\cap B)\]
Thus, the addition theorem on probability is proved.



Option ‘C’ is correct

Note: Here we may go wrong with the set theory formula. If the Venn diagram is used, then the formulae will be established easily. Then, by applying the probability to the sets, the required addition theorem on probability is obtained.