Assuming you have a container holding 4 red balls and 2 blue balls, what are the chances of selecting one red ball and one blue ball consecutively without returning the first pick?

Question

Anonymous · Accepted Answer

Assuming the events are independent, the probability of them not being the same
in 2 picks = P(RB) + P(BR)
P(RB) = P(R)*P(B) = 4/6 * 2/6 = 8/36 = 0.22
P(BR) = P(B)*P(R) = 0.22
P(not the same in 2 picks) = 0.22 + 0.22 = 0.44

Anonymous · Answer

* There are 6 balls in total, 4 Red and 2 Blue: Probability of picking one red
   and one blue without replacement is: 4/6*2/5 = 8/30 = 4/15

Assuming you have a container holding 4 red balls and 2 blue balls, what are the chances of selecting one red ball and one blue ball consecutively without returning the first pick?

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