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There are 55 balls in the box, of which are 25 red and the rest is black. We draw randomly 8 balls from the box. What is the probability that among the selected balls will be a) exactly 2 black balls, b) at least 2 black, c) no more than two black.

Problem a): probability of drawing 2 black balls and 6 red balls $$P(A)=\frac{\binom{30}2\binom{25}6}{\binom{55}8}\;\dot =\;6.33\%$$

Problem c): sum of probabilities of drawing 0, 1 or 2 black balls $$P(C)=\frac{\binom{25}8}{\binom{55}8}+\frac{\binom{30}1\binom{25}7}{\binom{55}8}+P(A)\;\dot =\;7.6\%$$

Problem b): the complement of problem b) $$P(B)=1-P(C)\;\dot =\;92.4\%$$

Please, did I understand and solve the problem correctly and in the best possible way? Thank you!

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