Total number of balls: 4+6+2 = 12
Let W1 shows the event that first ball is white, R1 shows the event that first is red and B1 shows the event that first is black. So we have
P(W1) = 4/12
P(B1) = 6/12
P(R1) = 2/12
Let B2 shows the event that second ball is black. After drawing first ball number of balls remaining is 11. So,
P(B2|W1) = 6/11
P(B2|B1) = 5/11
P(B2|R1) = 6/11
By the Baye's theorem, the probability of getting first ball white given that second ball is black is
P(W1 | B2) = [P(B2 | W1) P(W1) ] / [P(B2 | W1) P(W1)+P(B2 | B1) P(B1)+P(B2 | R1) P(R1)] = [ (4/11) * (4/12) ] / [(6/11) * (4/12) + (5/11) * (6/12) + (6/11) * (2/12)] = 16 / [24 +30 +12] = 0.2424
Answer: 0.2424
BONUS PROBLEM. (5 points) Consider the following experiment: two balls are pulled without replacement, one after...
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