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The first basket contains 4 blue and 5 red marbles; the

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The first basket contains 4 blue and 5 red marbles; the [#permalink]

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02 Sep 2006, 10:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The first basket contains 4 blue and 5 red marbles; the second basket contains 3 blue and 4 red marbles. One marble is randomly extracted from the first basket and put into the second. After that, a marble is extracted from the second basket. What is the probability that this marble is blue?

1. 1/3
2. 15/36
3. 31/72
4. 4/9
5. 11/18
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Re: Prob: Dependent probability - have I lost my marbles? [#permalink]

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02 Sep 2006, 20:11
Futuristic wrote:
The first basket contains 4 blue and 5 red marbles; the second basket contains 3 blue and 4 red marbles. One marble is randomly extracted from the first basket and put into the second. After that, a marble is extracted from the second basket. What is the probability that this marble is blue?

1. 1/3
2. 15/36
3. 31/72
4. 4/9
5. 11/18

Required probability = P(first pick was blue) * P(2nd pick was blue) + P(first pick was red)*P(2nd pick was blue)
= 4/9 * 4/8 + 5/9 * 3/8 = 31/72.

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03 Sep 2006, 14:07
Probability first blue/second blue = 2/9 (4/9*1/2)..1
Probability first red/second blue = 5/24 (5/9*3/8)..2

Probability second blue =1+2 = 31/72
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03 Sep 2006, 19:23
gk3.14 wrote:
Probability first blue/second blue = 2/9 (4/9*1/2)..1
Probability first red/second blue = 5/24 (5/9*3/8)..2

Probability second blue =1+2 = 31/72

Probability that first one is blue, 4c1/9c1 * 4/8 ---------1

Probability that frist one is red, 5/9*3/8------------2

1+2= 31/72
03 Sep 2006, 19:23
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