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

Anonymous · Answer

If the goal is to select specifically red, then blue, the likelihood is p(red)
*p(blue | one red selected) = (4/6)*(2/5) = (2/3)*(2/5) =0.267 . However, if the
question is around the likelihood of selected two different colored balls
consecutively, regardless of order, this is 0.267 + p(blue) p(red | one blue
selected) . The latter part of the statement is (1/3)*(4/5) = 0.267. The answer
then is 53%.

Anonymous · Answer

P(R) = 4/6 P(B|R) = 2/5 ... P both = 8/30, vice versa it's same thing 2/6 and
4/5 ... given that the two events can't both occur, total = 16/30. we can double
check this another way: p of 2r's is 4*3/3 = 12/30 and of two blues is 2*1/30.
those are the only ways we couldn't get mixed balls, so 30/30 - 14/30 = 16/30.

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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