From the first 100 natural numbers, two numbers a and b (with a first, then b) are selecte…

Question

From the first 100 natural numbers, two numbers $a$ and $b$ (with $a$ first, then $b$) are selected randomly without replacement. If the probability that $a-b \ge 10$ is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_.

Answer: 311

Accepted Answer

Total number of ways to choose the ordered pair $(a,b)$ is $100 \times 99$.

For a fixed $b$, we need $a \ge b+10$ with $a \le 100$, so $a$ can take $100-(b+10)+1 = 91-b$ values, valid for $b = 1, 2, \dots, 90$.

$$\text{Favourable cases} = \sum_{b=1}^{90}(91-b) = 90+89+\cdots+1 = \frac{90\times 91}{2} = 4095$$

$$P(a-b\ge 10) = \frac{4095}{100\times 99} = \frac{4095}{9900} = \frac{91}{220}$$

Since $\gcd(91,220)=1$, $m = 91$, $n = 220$, so $m+n = 311$.

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