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> $4$ out of $20$ balls are black. The others are white. If I pick $2$ balls out of those $20$ randomly, what is the probability that at least one of them is white?
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I could do this using a different approach: subtracting the probability that both are black from $1$. But while trying the direct approach, I've encountered a problem. The $2$ approaches I've tried using provide different answers each.

$1$st approach: $$\begin{cases}n={20\choose 2}\\ m=4\cdot 16 + {16\choose 2}\end{cases}\implies P=\frac{m}{n}=\frac{92}{95}$$

$2$nd approach: $$P=\frac{4}{20}\cdot\frac{16}{20}+\frac{16}{20}\cdot\frac{15}{20}=0.76$$

I think you'll be able to understand my reasoning: the $m$ shows the amount of ways I can choose both a white ball and a black ball $+$ the amount of ways I can choose two white ones. The $2$nd approach uses similar reasoning. So where is the mistake? Thanks.