Question ID: #808
From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $a-b\ge10$ is $\frac{m}{n}$ where $\gcd(m,n)=1,$ then $m+n$ is equal to
Solution:
Total number of ways to select two numbers $a$ and $b$ from 100 without replacement (order matters):
$$ \text{Total Cases} = 100 \times 99 = 9900 $$
Favorable condition: $a – b \ge 10 \implies b \le a – 10$.
We iterate through possible values of $a$.
Since $b$ must be at least 1, $a – 10 \ge 1 \implies a \ge 11$.
If $a = 11$, $b \le 1 \implies b \in \{1\}$ (1 value).
If $a = 12$, $b \le 2 \implies b \in \{1, 2\}$ (2 values).
…
If $a = 100$, $b \le 90 \implies b \in \{1, 2, \dots, 90\}$ (90 values).
Total Favorable Cases:
$$ \text{Sum} = 1 + 2 + 3 + \dots + 90 $$
$$ \text{Sum} = \frac{90 \times 91}{2} = 45 \times 91 = 4095 $$
Probability $P$:
$$ P = \frac{4095}{9900} $$
Divide numerator and denominator by 45:
$$ 4095 \div 45 = 91 $$
$$ 9900 \div 45 = 220 $$
$$ P = \frac{91}{220} $$
Here $m = 91$ and $n = 220$.
Check GCD(91, 220): $91 = 7 \times 13$, $220 = 2^2 \times 5 \times 11$. No common factors.
We need $m + n$:
$$ m + n = 91 + 220 = 311 $$
Ans. 311
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