Set Theoretical Notations of Probability - Practice Questions & MCQ

The Addition Rule of Probability 

(Probability of the event ‘A or B’ )

If $A$ and $B$ are any two events defined on a sample space, then the probability of occurrence of at least one of the event $A$ and $B$ is $P(A \cup B)$ and it equals $P(A)+P(B)-P(A \cap B)$.
From the set theory, we know that

$
n(A \cup B)=n(A)+n(B)-n(A \cap B)
$
Divide by $n(S)$ both side

$
\begin{aligned}
& \frac{n(A \cup B)}{n(S)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B)
\end{aligned}
$

Special case

If $A$ and $B$ are disjoint sets, i.e., they are mutually exclusive events, then $A \cap B=\varphi$. Therefore $P(A \cap B)=P(\varphi)$ $=0$

Thus, for mutually exclusive events $A$ and $B$, we have $P(A \cup B)=P(A)+P(B)$

If $A$ and $B$ are two events, then

$
(A-B) \cap(A \cap B)=\varphi \quad \text { and } \quad A=(A-B) \cup(A \cap B)
$

$
\begin{aligned}
& \text { So, } P(A)=P(A-B)+P(A \cap B)-0 \\
& =P\left(A \cap B^{\prime}\right)+P(A \cap B)
\end{aligned}
$

or $P(A)-P(A \cap B)=P\left(A \cap B^{\prime}\right)=P(A-B)$

$
\left[\because A-B=A \cap B^{\prime}\right]
$
Similarly, $P(B)-P(A \cap B)=P\left(B \cap A^{\prime}\right)=P(B-A)$
If $A, B$ and $C$ are any three events in a sample space $S$, then

$
P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C
$

If $\mathrm{A}, \mathrm{B}$ and C are any three mutually exclusive events in a sample space S , then $P(A \cup B \cup C)=P(A)+P(B)+P(C)$

Probability of event ‘not A’ or Complementary Event

If E is the any event and $\mathrm{E}^{\prime}$ be the complement of the event E . Since, E and $\mathrm{E}^{\prime}$ are disjoint and exhaustive sets.
\begin{array}{ll} 
& \mathrm{E} \cup \mathrm{E}^{\prime}=\mathrm{S} \\
\therefore & \mathrm{n}\left(\mathrm{E} \cup \mathrm{E}^{\prime}\right)=\mathrm{n}(\mathrm{~S}) \\
\Rightarrow & \mathrm{n}(\mathrm{E})+\mathrm{n}\left(\mathrm{E}^{\prime}\right)=\mathrm{n}(\mathrm{~S}) \\
\Rightarrow & \frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}+\frac{\mathrm{n}\left(\mathrm{E}^{\prime}\right)}{\mathrm{n}(\mathrm{~S})}=1 \\
\Rightarrow & \mathrm{P}(\mathrm{E})+\mathrm{P}\left(\mathrm{E}^{\prime}\right)=1 \\
\Rightarrow & \mathrm{P}(\mathrm{E})=1-\mathrm{P}(\text { not } \mathrm{E})=1-\mathrm{P}\left(\mathrm{E}^{\prime}\right)
\end{array}