Question

A bag contains 8 blue balls and 5 black balls. 2 successive draws of 3 balls are made without replacement. find the probability that the first drawing will give 3 black balls and the second 3 blue balls.

Answer 1

Answer on Question #67012, Math / Statistics and Probability.

We will use the formula

$$P(A \cap B) = P(A)P(B \mid A).$$

$A$ is the event such that the first drawing will give 3 black balls.

$B$ is the event such that the second drawing will give 3 blue balls.

Then $P(A) = \frac{\binom{3}{5}}{\binom{3}{13}} = \frac{\frac{5!}{3! \cdot 2!}}{\frac{13!}{3! \cdot 10!}} = \frac{5! \cdot 10!}{2! \cdot 13!} = \frac{3 \cdot 4 \cdot 5}{11 \cdot 12 \cdot 13} = \frac{5}{11 \cdot 13}$,

$$P(B \mid A) = \frac{\binom{3}{8}}{\binom{3}{10}} = \frac{\frac{8!}{3! \cdot 5!}}{\frac{10!}{3! \cdot 7!}} = \frac{8! \cdot 7!}{5! \cdot 10!} = \frac{6 \cdot 7}{9 \cdot 10} = \frac{7}{3 \cdot 5}.$$

Therefore

$$P(A \cap B) = P(A)P(B \mid A) = \frac{5}{11 \cdot 13} \cdot \frac{7}{3 \cdot 5} = \frac{7}{3 \cdot 11 \cdot 13} = \frac{7}{429} = 0.01632.$$

Hence the probability that the first drawing will give 3 black balls and the second 3 blue balls is 0.0163.

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