Question

BONUS PROBLEM. (5 points) Consider the following experiment: two balls are pulled without replacement, one after another, from the box containing four white, six black, and two red balls. What is the probability that the 1st ball is white given that the 2nd ball is black? (Round your answer to 2 places after the decimal point). (Your solutions have to include computational fin Answer:

Answer 1

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