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Set Theoretical Notations of Probability is considered one of the most asked concept.
32 Questions around this concept.
For three events A, B, and C, P(Exactly one of A or B occurs) =P(Exactly one of B or C occurs)
=P(Exactly one of C or A occurs) and P(All three events occur simultaneously) . Then the probability that at least one of the events occurs, is :
A number x is chosen at random from the set {1,2,3,4........100} Define the event : A= the chosen number x satisfies
Then P(A) is :
Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight?
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In a class of 200 students, 125 students have taken a programming language course, 85 students have taken a data structures course, 65 students have taken a computer organization course, 50 students have taken both programming languages and data: structures, 35 students have taken both programming languages and computing organization, 30 students have taken both data structures and computer organization, 15 students have taken all the three courses. How many students have not taken any of the three courses?
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}
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