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Bunuel
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 29 Jan 2019, 03:02
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B
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70% (02:08) correct 30% (02:26) wrong based on 261 sessions

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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.

A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8
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B
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Schools: IIM (A) ISB '24
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 29 Jan 2019, 03:58
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Bunuel
Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.

A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8
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Probability of Last ball chosen white = Total White balls in Box 2 / Total balls in Box 2

Box 1: 3 red and 5 white balls
Box 2: 4 red and 2 white balls

Case 1: Ball chosen Box 1 is White and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball white in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (5/8)*(3/7)

Case 2: Ball chosen Box 1 is Red and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball Red in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (3/8)*(2/7)

Total Favourable Probability = Probability of case 1+ Probability of Case 2

i.e. Required Probability = (5/8)*(3/7) + (3/8)*(2/7) = (15+6)/56 = 21/56 = 3/8

Answer: Option B
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 29 Jan 2019, 04:09
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Bunuel
Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.

A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8
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we can solve it considering two cases

case 1 : only white balls are chosen from box 1 and is transferred to box 2
P ( white ) ;5/8 * 3/7

case 2: red ball is chosen from box 1 and is transferred to box 2
P (white ) ; 3/8 * 2/7
5/8 * 3/7 + 3/8 * 2/7

15+6/56 = 21/56 ; 3/8
IMO B
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 22 Feb 2025, 18:22
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How is probability of choosing white ball from box 2 is 3/7 in the case 1?
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Bunuel
Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.

A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8

Probability of Last ball chosen white = Total White balls in Box 2 / Total balls in Box 2

Box 1: 3 red and 5 white balls
Box 2: 4 red and 2 white balls

Case 1: Ball chosen Box 1 is White and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball white in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (5/8)*(3/7)

Case 2: Ball chosen Box 1 is Red and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball Red in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (3/8)*(2/7)

Total Favourable Probability = Probability of case 1+ Probability of Case 2

i.e. Required Probability = (5/8)*(3/7) + (3/8)*(2/7) = (15+6)/56 = 21/56 = 3/8

Answer: Option B
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 23 Feb 2025, 00:50
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basanti7
How is probability of choosing white ball from box 2 is 3/7 in the case 1?
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Bunuel
Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 2 white balls. A ball is chosen at random from the first box and placed in the second box without observing its color. Then the ball is drawn from the second box. Find the probability that it is white.

A. 2/7
B. 3/8
C. 3/7
D. 4/7
E. 5/8

Probability of Last ball chosen white = Total White balls in Box 2 / Total balls in Box 2

Box 1: 3 red and 5 white balls
Box 2: 4 red and 2 white balls

Case 1: Ball chosen Box 1 is White and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball white in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (5/8)*(3/7)

Case 2: Ball chosen Box 1 is Red and then Ball chosen from Box 2 is White
Favourable Probability = Probability of chosen ball Red in box 1 * Probability of chosen ball white from box 2

i.e. Favourable Probability = (3/8)*(2/7)

Total Favourable Probability = Probability of case 1+ Probability of Case 2

i.e. Required Probability = (5/8)*(3/7) + (3/8)*(2/7) = (15+6)/56 = 21/56 = 3/8

Answer: Option B
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Because a white ball is transferred from Box 1 to Box 2, Box 2 now has 3 white balls and a total of 7 balls.
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Box 1 contains 3 red and 5 white balls, while box 2 contains 4 red and 29 Jan 2026, 07:19
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Deconstructing the Question

This is a Total Probability problem involving two mutually exclusive scenarios based on the color of the transferred ball.

Initial State:
Box 1: 3 Red, 5 White (Total 8)
Box 2: 4 Red, 2 White (Total 6)

Scenario 1: Transferred ball is RED
  • Probability of picking Red from Box 1: \(3/8\)
  • New Box 2 composition: 5 Red, 2 White (Total 7)
  • Probability of picking White from new Box 2: \(2/7\)
  • Combined Probability: \((3/8) * (2/7) = \) 6/56

Scenario 2: Transferred ball is WHITE
  • Probability of picking White from Box 1: \(5/8\)
  • New Box 2 composition: 4 Red, 3 White (Total 7)
  • Probability of picking White from new Box 2: \(3/7\)
  • Combined Probability: \((5/8) * (3/7) = \) 15/56

Total Probability
Sum the probabilities of both scenarios:
\(P(Total) = 6/56 + 15/56 = 21/56\)

Simplify by dividing by 7:
\(21/56 = \) 3/8

The correct answer is (B) .
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