Question

If an integer is chosen at random from the first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$ is:
(A) $\dfrac{{41}}{{100}}$
(B) $\dfrac{{33}}{{100}}$
(C) $\dfrac{1}{{10}}$
(D) None of these

Answer 1

Hint: The given question involves the concepts of probability and arithmetic progressions. We are to find the probability of choosing multiples of $4$ or $6$. We first find the number of multiples of $4$ and $6$ separately in the first $100$ positive integers. Then, we subtract the number of multiples of $12$ 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:
So, we have first $100$ positive integers.
Firstly, we find the number of multiples of $4$ in the first $100$ positive integers.
So, we get the series as: $4$, $8$, $12$, …… $100$.
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, the first term $ = a = 4$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 8 - 4 = 4\]
So, $d = 4$ .
We know the formula for the general term in an arithmetic progression is ${a_n} = a + \left( {n - 1} \right)d$.
So, considering $100$ as the nth term of the AP, we can find the number of terms in the AP that corresponds to $100$.
So, \[{a_n} = a + \left( {n - 1} \right)d = 100\]
Substituting the values of a and d in the above equation, we get,
$ \Rightarrow 4 + \left( {n - 1} \right)4 = 100$
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 4\left( {n - 1} \right) = 96$
Carrying out the calculations and dividing both sides of the equation by $4$, we get,
$ \Rightarrow \left( {n - 1} \right) = \dfrac{{96}}{4}$
$ \Rightarrow \left( {n - 1} \right) = 24$
Adding $1$ to both sides of the equation, we get,
$ \Rightarrow n = 25$
So, $100$ is the ${25^{th}}$ term of the sequence $4$, $8$, $12$, …… $100$.
Now, we find the number of multiples of $6$ in the first $100$ positive integers.
So, we get the series as: $6$, $12$, $18$, …… $96$.
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 = 6$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 12 - 6 = 6\]
So, $d = 6$ .
We know the formula for the general term in an arithmetic progression is ${a_n} = a + \left( {n - 1} \right)d$.
So, considering $96$ as the nth term of the AP, we can find the number of terms in the AP that corresponds to $96$.
So, \[{a_n} = a + \left( {n - 1} \right)d = 96\]
Substituting the values of a and d in the above equation, we get,
$ \Rightarrow 6 + \left( {n - 1} \right)6 = 96$
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 6\left( {n - 1} \right) = 96 - 6$
Carrying out the calculations and dividing both sides of the equation by $6$, we get,
$ \Rightarrow \left( {n - 1} \right) = \dfrac{{90}}{6}$
$ \Rightarrow \left( {n - 1} \right) = 15$
Adding $1$ to both sides of the equation, we get,
$ \Rightarrow n = 16$
So, $96$ is the ${16^{th}}$ term of the sequence $6$, $12$, $18$, …… $96$.
Now, we know that the least common multiple of $4$ and $6$ is $12$.
Now, we find the number of multiples of $12$ in the first $100$ positive integers.
So, we get the series as: $12$, $24$, …… $96$.
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 = 12$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 24 - 12 = 12\]
So, $d = 12$ .
We know the formula for the general term in an arithmetic progression is ${a_n} = a + \left( {n - 1} \right)d$.
So, considering $96$ as the nth term of the AP, we can find the number of terms in the AP that corresponds to $96$.
So, \[{a_n} = a + \left( {n - 1} \right)d = 96\]
Substituting the values of a and d in the above equation, we get,
$ \Rightarrow 12 + \left( {n - 1} \right)12 = 96$
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 12\left( {n - 1} \right) = 96 - 12$
Carrying out the calculations and dividing both sides of the equation by $12$, we get,
\[ \Rightarrow \left( {n - 1} \right) = \dfrac{{84}}{{12}}\]
\[ \Rightarrow \left( {n - 1} \right) = 7\]
Adding $1$ to both sides of the equation, we get,
$ \Rightarrow n = 8$
So, $96$ is the ${8^{th}}$ term of the sequence $12$, $24$, …… $96$.
So, the number of multiples of four or six can be calculated by adding the multiples of four and six and then subtracting the number of multiples of twelve from it.
So, the number of multiples of four or six will be equal to $\text{the number of multiples of four + number of multiples of six - the common multiples of four and six.}$
 $ = 25 + 16 - 8$
$ = 33$ (This is the number favourable outcomes to get the multiples of 4 or 6)
Therefore, the number of multiples of $4$ or $6$ is $33$. Hence, the probability of choosing a multiple of $4$ or $6$ from the first $100$ positive integers is $\dfrac{{\text{Number of favourable outcomes}}}{{\text{Total number of outcomes}}} = \dfrac{{33}}{{100}}$.
Hence, option (B) is correct.

Note:
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: ${a_n} = a + \left( {n - 1} \right)d$. We also must know the basic formula of probability to solve the given problem.