Suppose balls are randomly removed one by one without replacement from an urn that the urn initially contains 20 red, 10 blue, and 8 green balls, then which of the following option(s) is / are true?

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the probability that all of the red balls are removed before all of the blue ones have been removed is $1 / 3$

the probability that the colors are depleted in the order blue, red, green is 8/57

the probability that the group of blue balls is the first of the three groups to be removed is 64/171

the probability that the colors are depleted in the order blue, red, green is 2/3

answer is A, B, C.

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Detailed Solution

(a) This is the probability that the last red or blue ball to be removed is a blue ball. Because it is equally likely to be any of the 30 red or blue balls, the probability that it is blue is $1 / 3$ .
(b) Let $B_{1}, R_{2}, G_{3}$ denote, respectively, the events that the first color removed is blue, the second is red, and the third is green. Then
$P (B_{1} R_{2} G_{3}) = P (G_{3}) P (R_{2} ∣ G_{3}) P (B_{1} ∣ R_{2} G_{3}) = \frac{8}{38} \frac{20}{30} = \frac{8}{57}$
where $P (G_{3})$ is just the probability that the very last ball is green and $P (R_{2} ∣ G_{3})$ is computed by noting that, given that the last ball is green, each of the 20 red and 10 blue balls is equally likely to be the last of that group to be removed, so the probability that it is one of the red balls is 20/30. (Of course, $P (B_{1} ∣ R_{2} G_{3}) = 1$ )
(c) $P (B_{1}) = P (B_{1} G_{2} R_{3}) + P (B_{1} R_{2} G_{3}) = \frac{20}{38} \frac{8}{18} + \frac{8}{57} = \frac{64}{171}$