Probability Question - None

Question

For Problems 4 and 5: Let $U_1$ and $U_2$ be two urns such that $U_1$ contains 3 white and 2 red balls, and $U_2$ contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from $U_1$ and put into $U_2$. However, if tail appears, then 2 balls are drawn at random from $U_1$ and put into $U_2$. Now 1 ball is drawn at random from $U_2$.

Problem 4: The probability of the drawn ball from $U_2$ being white is

Accepted Answer

Step 1: Condition on coin toss outcome.

P(White from U₂) = P(White | Head)·P(Head) + P(White | Tail)·P(Tail)

= P(White | Head)·(1/2) + P(White | Tail)·(1/2)

Step 2 (Case Head - 1 ball transferred):

From U₁ (3W, 2R), draw 1 ball:

P(W transferred) = 3/5 → U₂ has 2W, 1R (total 3) → P(W drawn) = 2/3
P(R transferred) = 2/5 → U₂ has 1W, 1R (total 2) → P(W drawn) = 1/2

P(White | Head) = (3/5)·(2/3) + (2/5)·(1/2) = 2/5 + 1/5 = 3/5

Step 3 (Case Tail - 2 balls transferred):

From U₁ (3W, 2R), draw 2 balls:

P(2W) = C(3,2)/C(5,2) = 3/10 → U₂ has 3W, 1R (total 4) → P(W drawn) = 3/4
P(1W, 1R) = C(3,1)·C(2,1)/C(5,2) = 6/10 → U₂ has 2W, 2R (total 4) → P(W drawn) = 1/2
P(2R) = C(2,2)/C(5,2) = 1/10 → U₂ has 1W, 2R (total 3) → P(W drawn) = 1/3

P(White | Tail) = (3/10)·(3/4) + (6/10)·(1/2) + (1/10)·(1/3) = 9/40 + 12/40 + 4/120 = 27/120 + 36/120 + 4/120 = 67/120

Step 4: Combine results.

P(White from U₂) = (1/2)·(3/5) + (1/2)·(67/120) = 3/10 + 67/240 = 72/240 + 67/240 = 139/240

∴ Answer: B

Question:

Step-by-Step Solution

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