Discussion
A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is
Answer: D
The probability is 1/2.
Drawing one after another without replacement gives the same chances as drawing three together, so count combinations:
\[ \begin{aligned} P &= \frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \times 15}{120} \\ &= \frac{60}{120} = \frac{1}{2} \end{aligned} \]
The same answer comes from ordered draws: the chance of red, black, black in that order is \(\frac{4}{10}\cdot\frac{6}{9}\cdot\frac{5}{8} = \frac{1}{6}\), and the red ball can come in any of 3 positions, giving 1/2. Forgetting those 3 positions gives 1/6, which is not offered; 1/20, 1/12 and 3/10 do not match the count.
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