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

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Intern
Joined: 11 Oct 2008
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An urn contains 5 red and 5 black balls. A ball is drawn at [#permalink]

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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

Kudos [?]: [0], given: 0

VP
Joined: 05 Jul 2008
Posts: 1402

Kudos [?]: 437 [1], given: 1

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22 Oct 2008, 10:43
1
KUDOS
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

Kudos [?]: 437 [1], given: 1

SVP
Joined: 07 Nov 2007
Posts: 1792

Kudos [?]: 1058 [1], given: 5

Location: New York

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22 Oct 2008, 10:48
1
KUDOS
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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