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Hint: The given question uses the concepts of probability and divisibility. Recall the formula for finding the probability of choosing $x$ items from a group of $y$ items , where $x\le y$. Also, recall the test of divisibility by $8$.

Before proceeding with the solution , first, we should know about probability and divisibility.

Probability is defined as the likeness of an event to occur. Mathematically , it is defined as the ratio of number of favorable outcomes to the number of possible outcomes.

Divisibility rules are a quick way to determine whether large numbers are divisible by a given number.

Now , the divisibility rule for a number to be divisible by $8$ is “ if the last three digits of a number are divisible by $8$ , then the number is divisible by $8$ .”

We know, the numbers between $1$ and $100$ are two-digit numbers. So, we cannot apply the divisibility rule to test whether a number is divisible by $8$ or not. So, we have to pick numbers by adding $8$ to every multiple of $8$ . The first multiple is $8$ . The next will be $8+8=16$ . Similarly, the other numbers divisible by $8$ are $24,32,40,48,56,64,72,80,88$ and $96$ . So, the total number of numbers between $1$ and $100$ that are divisible by $8$ is $12$ .

Now, to find the value of $m$ , we first need to find the probability of choosing a number divisible by $8$ from a set of numbers between $1$ and $100$ .

The favorable event is choosing a number between $1$ and $100$ which is divisible by $8$.

So , the number of favorable outcomes is $12$.

The total possible outcome is choosing any number between $1$ and $100$. So , the number of possible outcomes is $100$ .

So , probability $P=\dfrac{12}{100}=\dfrac{3}{25}$.

Now , in the question , the value of the probability is given as $\dfrac{m}{25}$ .

So, $\dfrac{m}{25}=\dfrac{3}{25}$ .

Or, $m=3$ .

Hence, the value of $m$ is equal to $3$.

Note: To find the number of numbers between $1$ and $100$ that are divisible by $8$ , we can divide $100$ by $8$ , and the quotient will be our required value , i.e. $\dfrac{100}{8}=12(rem=4)$. So, the number of numbers between $1$ and $100$ that are divisible by $8$ is $12$.

Before proceeding with the solution , first, we should know about probability and divisibility.

Probability is defined as the likeness of an event to occur. Mathematically , it is defined as the ratio of number of favorable outcomes to the number of possible outcomes.

Divisibility rules are a quick way to determine whether large numbers are divisible by a given number.

Now , the divisibility rule for a number to be divisible by $8$ is “ if the last three digits of a number are divisible by $8$ , then the number is divisible by $8$ .”

We know, the numbers between $1$ and $100$ are two-digit numbers. So, we cannot apply the divisibility rule to test whether a number is divisible by $8$ or not. So, we have to pick numbers by adding $8$ to every multiple of $8$ . The first multiple is $8$ . The next will be $8+8=16$ . Similarly, the other numbers divisible by $8$ are $24,32,40,48,56,64,72,80,88$ and $96$ . So, the total number of numbers between $1$ and $100$ that are divisible by $8$ is $12$ .

Now, to find the value of $m$ , we first need to find the probability of choosing a number divisible by $8$ from a set of numbers between $1$ and $100$ .

The favorable event is choosing a number between $1$ and $100$ which is divisible by $8$.

So , the number of favorable outcomes is $12$.

The total possible outcome is choosing any number between $1$ and $100$. So , the number of possible outcomes is $100$ .

So , probability $P=\dfrac{12}{100}=\dfrac{3}{25}$.

Now , in the question , the value of the probability is given as $\dfrac{m}{25}$ .

So, $\dfrac{m}{25}=\dfrac{3}{25}$ .

Or, $m=3$ .

Hence, the value of $m$ is equal to $3$.

Note: To find the number of numbers between $1$ and $100$ that are divisible by $8$ , we can divide $100$ by $8$ , and the quotient will be our required value , i.e. $\dfrac{100}{8}=12(rem=4)$. So, the number of numbers between $1$ and $100$ that are divisible by $8$ is $12$.

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