Question

A natural number is chosen at random from amongst the first $300$. What is the probability that the number chosen is a multiple of $2\mathrm{or}3\mathrm{or}5$ ?

• A. $\frac{1}{10}$
• B. $\frac{11}{15}$
• C. $\frac{4}{150}$
• D. $\frac{17}{30}$

Answer 1

The correct option is B

$\frac{11}{15}$

Explanation for the correct option:
Finding $n\left(S\right),n\left(A\right),n\left(B\right),\mathrm{and}n\left(C\right)$

Let $S$ be the number of ways in which a natural number can be chosen from amongst the first $300$.

$n\left(S\right)={}^{300}C_1=300$

Let $A$ be the event that the chosen number is multiple of $2$. All even numbers are divided by $2$,

$\Rightarrow n\left(A\right)={}^{150}C_1=150$

Let $B$ be the event that the chosen number is multiple of $3$. In first $300$, $100$ numbers are divisible by $3$,

$\Rightarrow n\left(B\right)={}^{100}C_1=100$

Let $C$ be the event that the chosen number is multiple of $5$. In first $300$, $60$ numbers are divisible by $5$,

$\Rightarrow n\left(C\right)={}^{60}C_1=60$

Finding $n\left(A\cap B\right),n\left(B\cap C\right),n\left(A\cap C\right),\mathrm{and}n\left(A\cap B\cap C\right)$.

The chosen number is multiple of $2$ and $3$, it means the chosen number is multiple of $6$,

$\Rightarrow n\left(A\cap B\right)={}^{50}C_1=50$

The chosen number is multiple of $3$ and $5$, it means the chosen number is multiple of $15$,

$\Rightarrow n\left(B\cap C\right)={}^{20}C_1=20$

The chosen number is multiple of $2$ and $5$, it means the chosen number is multiple of $10$

$\Rightarrow n\left(A\cap C\right)={}^{30}C_1=30$

The chosen number is multiple of $2,3$ and $5$, it means the chosen number is multiple of $30$

$\Rightarrow n\left(A\cap B\cap C\right)={}^{10}C_1=10$

Finding the required probability.

The required probability $=P\left(A\cup B\cup C\right)$

$\begin{array}{rcl}P\left(A\cup B\cup C\right)& =& P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)-P\left(A\cap B\cap C\right)\\ & =& \frac{n\left(A\right)}{n\left(S\right)}+\frac{n\left(B\right)}{n\left(S\right)}+\frac{n\left(C\right)}{n\left(S\right)}-\frac{n\left(A\cap B\right)}{n\left(S\right)}-\frac{n\left(B\cap C\right)}{n\left(S\right)}-\frac{n\left(A\cap C\right)}{n\left(S\right)}+\frac{n\left(A\cap B\cap C\right)}{n\left(S\right)}\\ & =& \frac{150}{300}+\frac{100}{300}+\frac{60}{300}-\frac{50}{300}-\frac{20}{300}-\frac{30}{300}+\frac{10}{300}\\ & =& \frac{220}{300}=\frac{11}{15}\end{array}$

Hence, the correct answer is option B.