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 5: Given that the drawn ball from $U_2$ is white, the probability that head appeared on the coin is

Accepted Answer

Step 1: Set up initial conditions

U₁: 3 white, 2 red (total 5 balls)
U₂: 1 white (initially)
P(Head) = P(Tail) = 1/2

Step 2: Find P(White from U₂ | Head)

If Head: Draw 1 ball from U₁ and put into U₂
• P(1 white drawn) = 3/5 → U₂ has 2 white, 0 red → P(white from U₂) = 2/2 = 1
• P(1 red drawn) = 2/5 → U₂ has 1 white, 1 red → P(white from U₂) = 1/2

P(White | Head) = (3/5)(1) + (2/5)(1/2) = 3/5 + 1/5 = 4/5

Step 3: Find P(White from U₂ | Tail)

If Tail: Draw 2 balls from U₁ and put into U₂
• P(2 white drawn) = C(3,2)/C(5,2) = 3/10 → U₂ has 3 white, 0 red → P(white) = 1
• P(1 white, 1 red) = C(3,1)·C(2,1)/C(5,2) = 6/10 → U₂ has 2 white, 1 red → P(white) = 2/3
• P(2 red drawn) = C(2,2)/C(5,2) = 1/10 → U₂ has 1 white, 2 red → P(white) = 1/3

P(White | Tail) = (3/10)(1) + (6/10)(2/3) + (1/10)(1/3) = 3/10 + 4/10 + 1/30 = 9/30 + 12/30 + 1/30 = 22/30 = 11/15

Step 4: Apply Bayes' Theorem

P(Head | White) = [P(White | Head)·P(Head)] / [P(White | Head)·P(Head) + P(White | Tail)·P(Tail)]

= [(4/5)(1/2)] / [(4/5)(1/2) + (11/15)(1/2)]

= [4/10] / [4/10 + 11/30]

= [12/30] / [12/30 + 11/30]

= 12/23

∴ Answer: D

Question:

Step-by-Step Solution

Master Probability with Mathbee