Question

If \[{{A}_{1}},{{A}_{2}},{{A}_{3}},.....{{A}_{n}}\] are any $n$ events, then
A. \[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}})=P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})+...+P({{A}_{n}})\]
B. \[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}})>P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})+...+P({{A}_{n}})\]
C. \[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}})\le P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})+...+P({{A}_{n}})\]
D. None of these

Answer 1

Hint: In this question, we are to find the probability of union $n$ events. By using set theory, the required probability of $n$ events are calculated. The principle of mathematical induction is applied for finding this probability.



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:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
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:
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
The principle of mathematical induction is a process/technique used to prove that the given statement is true for the natural numbers.



Complete step by step solution:Consider $n$ events \[{{A}_{1}},{{A}_{2}},{{A}_{3}},.....{{A}_{n}}\].
According to set theory,
\[{{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}}\le S\text{ }...\text{(1)}\]
Where S is the sample space
Applying probability to (1),
\[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}})\le P(S)\text{ }...\text{(2)}\]
From the definition of probability, we can write
$A\subseteq S$ and $P(A)\ge 0$
I.e., $P({{A}_{1}})\ge 0,P({{A}_{2}})\ge 0,...P({{A}_{n}})\ge 0$
Then,
\[\begin{align}
  & P(S)=\sum\limits_{a\in S}{P(a)} \\
 & \Rightarrow P(S)=P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})+...+P({{A}_{n}})...(3) \\
\end{align}\]
Since \[{{A}_{1}},{{A}_{2}},{{A}_{3}},.....{{A}_{n}}\in S\]
Then, from (2) and (3) we get,
\[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}\cup .....\cup {{A}_{n}})\le P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})+...+P({{A}_{n}})\]



Option ‘C’ is correct

Note: Here we may go wrong with the set theory formula. Here every set is the subset of the sample. So, the probability of the union of all the sets is less than the sum of their individual probabilities in a sample. Here we can also use the mathematical induction technique to achieve the proof.