From the first 100 natural numbers, two numbers first a and then b are selected randomly w…

Question

From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $\mathrm{a}-\mathrm{b} \geqslant 10$

Accepted Answer

Let $b=1$

$\Rightarrow a>11\Rightarrow$ a can take 90 values from ${11,12,\ldots 100}$

Let $b=2\Rightarrow a>12\Rightarrow 8>$ values and so on till $b=99\Rightarrow$ a can only take $a=100$, only 1 value \Rightarrow  favourable cases $=1+2+3+\ldots .+90$

$=\frac{\frac{90	imes 91}{2}}{100	imes 99}=\frac{91}{220}\Rightarrow m+n=311$

Solutions · 1