Probability Calculation for Random Selection

The problem asks for the value of $m + n$, where the probability of selecting two numbers $a$ and $b$ randomly without replacement from the first 100 natural numbers such that $a - b \ge 10$ is $\frac{m}{n}$, and $\gcd(m, n) = 1$. The numbers are selected sequentially, meaning the order matters.

Determining Total Possible Outcomes

The first number, $a$, can be any of the 100 natural numbers. Since the selection is without replacement, the second number, $b$, can be any of the remaining 99 numbers.

Total number of possible ordered pairs $(a, b)$ is:

$ \text{Total Outcomes} = 100 \times 99 = 9900 $

Calculating Favorable Outcomes

We need the number of pairs $(a, b)$ such that $a - b \ge 10$, or equivalently, $a \ge b + 10$. We can count these by iterating through possible values of $b$:

• If $b = 1$, $a$ must be $11, 12, \dots, 100$. Number of values for $a$ is $100 - 11 + 1 = 90$.
• If $b = 2$, $a$ must be $12, 13, \dots, 100$. Number of values for $a$ is $100 - 12 + 1 = 89$.
• ...
• If $b = 90$, $a$ must be $100$. Number of values for $a$ is $1$.
• If $b \ge 91$, $a \ge b + 10 \ge 101$, which is impossible since $a \le 100$.

The total number of favorable outcomes is the sum of the number of values for $a$ for each possible $b$:

$ \text{Favorable Outcomes} = 90 + 89 + 88 + \dots + 1 $

This is the sum of the first 90 natural numbers, which can be calculated using the formula for the sum of an arithmetic series, $S_n = \frac{n(n+1)}{2}$:

$ \text{Favorable Outcomes} = \frac{90 \times (90 + 1)}{2} = \frac{90 \times 91}{2} = 45 \times 91 = 4095 $

Probability Calculation and Simplification

The probability $P$ is the ratio of favorable outcomes to total outcomes:

$ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{4095}{9900} $

Now, we simplify the fraction $\frac{4095}{9900}$. Both numerator and denominator are divisible by 5:

$ \frac{4095 \div 5}{9900 \div 5} = \frac{819}{1980} $

Both are divisible by 9:

$ \frac{819 \div 9}{1980 \div 9} = \frac{91}{220} $

The prime factorization of 91 is $7 \times 13$. The prime factorization of 220 is $2^2 \times 5 \times 11$. Since there are no common factors, the fraction $\frac{91}{220}$ is in its simplest form.

Therefore, $m = 91$ and $n = 220$. We verify that $\gcd(91, 220) = 1$.

Final Calculation of m + n

We need to find the value of $m + n$:

$ m + n = 91 + 220 = 311 $