Question
A number is randomly selected from the first $50$ natural numbers. Find the probability that it is a multiple of $2$ or $3 .$

Answer

If one number is randomly selected from the first $50$ natural numbers then the number of mutually exclusive, exhaustive and equi-probable outcomes in the sample space of this random experiment will be $n={ }^{50} C_{1}=50$.
If event $A$ denotes that the number selected is a multiple of $2$ and event $B$ denotes that the number selected is a multiple of $3$ then the event that the selected number is a multiple of $2$ or $3$ will be denoted by $A \cup B. ($This event can also be denoted as $B \cup A$. According to set theory, $A \cup B=B \cup A ).$
To find the probability of $A \cup B$, the union of events $A$ and $B$ by the law of addition of probability, we will first find $P(A), P(B)$ and $P(A \cap B)$.
$A=$ Event that the selected number is a multiple of $2 =\{2,4,6, \ldots, 50\}$
Hence, the number of favourable outcomes of event $A$ will be $m=25$.
Probability of event $A P(A)=\frac{m}{n}$
$ =\frac{25}{50} $
$B=$ Event that the selected number is a multiple of $3 =\{3,6,9, \ldots, 48\}$ IIence,
the number of favourable outcomes of event $B$ will be $m=16$.
Probability of event $B P(B)=\frac{m}{n}$ $ =\frac{16}{50} $
$A \cap B=$ Event that the selected number is a multiple of $2$ and $3$ that is multiple the $LCM$ of $2$ and $3$ which is $6 .$
$=\{6,12,18, \ldots, 48\}$
Hence, the number of favourable outcomes of event $A \cap B$ will be $m=8$.
Probability of event $A \cap B P(A \cap B)=\frac{m}{n}$ $ =\frac{8}{50} $
From the law of addition of probability, $ P(A \cup B) =P(A)+P(B)-P(A \cap B)$
$ =\frac{25}{50}+\frac{16}{50}-\frac{8}{50}$
$ =\frac{25+16-8}{50}$
$ =\frac{33}{50} $
Required probability $=\frac{33}{50}$

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