Question

Twelve tickets are numbered from $1$ to $12$. One ticket is drawn at random, then the probability of the number to be divisible by $2$ or $3$ is:
(A) $\dfrac{2}{3}$
(B) $\dfrac{7}{{12}}$
(C) $\dfrac{5}{6}$
(D) $\dfrac{3}{4}$

Answer 1

The given question involves the concepts of probability and divisibility. We are to find the probability of choosing numbers divisible by $2$ or $3$ from twelve tickets numbered from $1$ to $12$. We first find the number of multiples of $2$ or $3$ separately in the twelve tickets. Then, we subtract the number of multiples of $6$ from the sum of the number of multiples of $4$ and $6$. For finding out the number of terms that correspond to a given term in an arithmetic progression, we must know the formula for a general term in an AP: ${a_n} = a + \left( {n - 1} \right)d$.

Complete step by step answer:
We have twelve tickets numbered from $1$ to $12$.
Firstly, we find the number of multiples of $2$ in the numbers from $1$ to $12$.
So, we get the series as: $2$, $4$, $6$, …… $12$.
We can notice that the difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression.
Here, first term $= a = 2$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio $= d = 4 - 2 = 2$
So, $d = 2$ .
We know the formula for the general term in an arithmetic progression is ${a_n} = a + \left( {n - 1} \right)d$.
So, considering $12$ as the nth term of the AP, we can find the number of term in the AP that corresponds to $12$.
So, ${a_n} = a + \left( {n - 1} \right)d = 12$
Substituting the values of a and d in the above equation, we get,
$\Rightarrow 2 + \left( {n - 1} \right)\left( 2 \right) = 12$
Now, in order to solve the above equation for the value of n, we shift all the constant terms to the right side of the equation. So, we get,
$\Rightarrow \left( {n - 1} \right)\left( 2 \right) = 12 - 2$
Carrying out the calculations and dividing both sides of the equation by $2$, we get,
$\Rightarrow \left( {n - 1} \right) = \dfrac{{10}}{2}$
$\Rightarrow \left( {n - 1} \right) = 5$
Adding $1$ to both sides of the equation, we get,
$\Rightarrow n = 6$
So, there are $6$ numbers divisible by $2$.
Now, we find the number of multiples of $3$ from $1$ to $12$.
So, we get the series as: $3$, $6$, ….. $12$.
We can notice that the difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression.
Here, first term $= a = 3$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio $= d = 6 - 3 = 3$
So, $d = 3$ .
We know the formula for the general term in an arithmetic progression is ${a_n} = a + \left( {n - 1} \right)d$.
So, considering $12$ as the nth term of the AP, we can find the number of term in the AP that corresponds to $12$.
So, ${a_n} = a + \left( {n - 1} \right)d = 12$
Substituting the values of a and d in the above equation, we get,
$\Rightarrow 3 + 3\left( {n - 1} \right) = 12$
Now, in order to solve the above equation for the value of n, we shift all the constant terms to the right side of the equation. So, we get,
$\Rightarrow 3\left( {n - 1} \right) = 12 - 3$
Carrying out the calculations and dividing both sides of the equation by $3$, we get,
$\Rightarrow \left( {n - 1} \right) = \dfrac{9}{3}$
$\Rightarrow \left( {n - 1} \right) = 3$
Adding $1$ to both sides of the equation, we get,
$\Rightarrow n = 4$
So, there are $4$ numbers divisible by $2$.
Now, we know that the least common multiple of $2$ and $3$ is $6$.
Now, we find the number of multiples of $6$ from $1$ to $12$.
So, we get the series as: $6$, $12$.
So, there are clearly only two numbers that are divisible by $6$.
So, the number of multiples of two or three can be calculated by adding the multiples of two and three and then subtracting the number of multiples of three from it.
So, the number of multiples of two or three $= 6 + 4 - 2$
$= 8$
Therefore, the number of multiples of $2$ or $3$ is $8$. Hence, the probability of choosing a multiple of $2$ or $3$ from $1$ to $12$ is $\dfrac{{\text{Number of favourable outcomes}}}{{\text{Total number of outcomes}}} = \dfrac{8}{{12}} = \dfrac{3}{4}$. Hence, the option (D) is correct.

Note:
The basic concepts and formulae of probability should be known before attempting such questions. Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. Any term of an arithmetic progression can be calculated if we know the first term and the common difference of the arithmetic series as: ${a_n} = a + \left( {n - 1} \right)d$. Calculations must be verified once to be sure of the final answer.