Question

We have 3 boxes. Box A contains 1 yellow ball and 2 red balls. Box B conatins 2 yellow balls and 2 red balls. Box C contains 2 yellow balls and 1 red ball. First, we randomly choose between Box A and Box B. Then we randomly choose a ball from the selected box and put the ball into Box C. Lastly, we randomly choose a ball from Box C.

Let E1 be the event that Box A is chosen first. Let E2 be the event that the first ball chosen is red. Let E3 be the event that the last ball chosen is red.

Find P(E2/E3)

Answer 1

We first compute here all the outcomes when E3 is occurring:

• E1, E2 and E3: First box chosen is A, the ball from first box chosen is red, and the last ball chosen is red.
  Prob. = 0.5*(2/3)*0.5 = 1/6
• not E1, E2 and E3: First box chosen is B, the ball from first box chosen is red, and the last ball chosen is red.
  Prob. = 0.5*(2/4)*0.5 = 1/8
• E1, not E2 and E3: First box chosen is A, the ball from first box chosen is not red, and the last ball chosen is red.
  Prob. = 0.5*(1/3)*(1/4) = 1/24
• not E1, not E2 and E3: First box chosen is B, the ball from first box chosen is not red, and the last ball chosen is red.
  Prob. = 0.5*(2/4)*(1/4) = 1/16

Therefore, we have here:
P(E3) = (1/6) + (1/8) + (1/24) + (1/16) = (8 + 6 + 2 + 3)/48 = 19/48

P(E2 and E3) = (1/6) + (1/8) = 7/24

Therefore the conditional probability now is computed using Bayes theorem here as:
P(E2 | E3) = P(E2 and E3) / P(E3) = (7/24) / (19/48) = 14/19

Therefore 14/19 = 0.7368 is the required probability here.