Discussion
A box contains 4 red and 6 white marbles. Two successive draws of 3 balls are made without replacement. The probability that in the first draw, all the 3 balls are white and in the second draw, all the 3 balls are red, is
Answer: A
The probability is \(\frac{2}{105}\).
First draw, 3 white from 6 white among 10 balls; second draw, 3 red from the 4 red among the 7 that remain:
\[ \begin{aligned} P &= \frac{{}^{6}C_{3}}{{}^{10}C_{3}} \times \frac{{}^{4}C_{3}}{{}^{7}C_{3}} \\ &= \frac{20}{120} \times \frac{4}{35} = \frac{4}{210} = \frac{2}{105} \end{aligned} \]
Forgetting that the first draw leaves only 7 balls (using \({}^{10}C_{3}\) twice) gives \(\frac{1}{6} \times \frac{1}{30} = \frac{1}{180}\); \(\frac{4}{105}\) doubles the answer.
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