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An urn contains 5 red and 5 black balls. A ball is drawn at

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An urn contains 5 red and 5 black balls. A ball is drawn at [#permalink]

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New post 22 Oct 2008, 09:51
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An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is
noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random. What is the probability that
the second ball is red?
Ans: 1/2

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Joined: 05 Jul 2008
Posts: 1402

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Re: probability [#permalink]

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New post 22 Oct 2008, 10:43
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First ball red - 5/10 = 1/2

Add 2 more red balls. toal balls = 12 red =7 probability of taking a red ball = 7/12

So probability of both first and second red balls = 1/2 X 7/12

First ball black is 5/10 = 1/2

Add 2 more black balls, total = 12, red =5 probability of taking a red ball = 5/12

So probability of both first and second red balls = 1/2 X 5/12

combined probability = 1/2 X 5/12 + 1/2 X 7/12 = 1/2 (5/12 + 7/12 ) = 1/2

PS: Please do not post answers along with Q's

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Re: probability [#permalink]

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New post 22 Oct 2008, 10:48
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earthwork wrote:
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is
noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random. What is the probability that
the second ball is red?
Ans: 1/2



= prob: first ball is red and secon ball is red + prob first ball is black and second ball is red
= p(RR) + p(BR)
=5/10 * 7/12+ 5/10 *5/12
= 1/2
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Kudos [?]: 1058 [1], given: 5

Re: probability   [#permalink] 22 Oct 2008, 10:48
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An urn contains 5 red and 5 black balls. A ball is drawn at

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