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There are 3 black balls and 10 white balls. If one is to pick 5 balls,

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Intern
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Joined: 06 Nov 2011
Posts: 36

Kudos [?]: 8 [0], given: 50

Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
GPA: 3
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There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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New post 21 Nov 2011, 10:27
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I am struggling with this question to solve (Source:its from a math book)

There are 3 black balls and 10 white balls. If one is to pick 5 balls, What is the probability of picking 2 black and 3 white?

Answer I got:
[Reveal] Spoiler:
\(\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%\)

Is there another approach?

Kudos [?]: 8 [0], given: 50

Manager
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Joined: 31 May 2011
Posts: 85

Kudos [?]: 67 [0], given: 4

Location: India
Concentration: Finance, International Business
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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New post 21 Nov 2011, 11:14
M3tm4n wrote:
I am struggling with this question to solve (Source:its from a math book)

There are 3 black balls and 10 white balls.
if one is to pick 5 balls, What is the probability of picking 2 black and 3 white?

Answer I got:
[Reveal] Spoiler:
\(\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%\)

Is there another approach?


the denominator should be 13C5 instead of 15C5

Kudos [?]: 67 [0], given: 4

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Veritas Prep GMAT Instructor
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Joined: 16 Oct 2010
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Kudos [?]: 18108 [1], given: 236

Location: Pune, India
Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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New post 21 Nov 2011, 22:48
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Expert's post
M3tm4n wrote:
I am struggling with this question to solve (Source:its from a math book)

There are 3 black balls and 10 white balls.
if one is to pick 5 balls, What is the probability of picking 2 black and 3 white?

Answer I got:
[Reveal] Spoiler:
\(\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%\)

Is there another approach?


Picking simultaneously is same as picking one after another.

So, let's see the probability of picking BBWWW.
(3/13)*(2/12)*(10/11)*(9/10)*(8/9)

But there are many other ways possible e.g. BWWWB, WWBBW etc
Number of all such arrangements = 5!/(2!*3!)

Required answer = (3/13)*(2/12)*(10/11)*(9/10)*(8/9)*5!/(2!*3!) = 40/143
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Kudos [?]: 18108 [1], given: 236

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Manager
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Joined: 12 Jul 2011
Posts: 139

Kudos [?]: 53 [2], given: 42

Concentration: Operations, Strategy
GMAT 1: 680 Q46 V37
WE: Engineering (Telecommunications)
Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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New post 22 Nov 2011, 03:13
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Probability of picking 5 balls from 10+3 Balls is 13C5
Probability of picking 2 Black balls from 3 is 3C2
Probability of picking 3 white balls from 10 is 10C3

So finally if one has to pick 5 balls, probability of picking 2 black and 3 white from 13 balls is:
( 3C2 * 10C3 ) / 13C5 = 40/143 :)

Kudos [?]: 53 [2], given: 42

Manager
Manager
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Joined: 29 Oct 2011
Posts: 180

Kudos [?]: 164 [0], given: 19

Concentration: General Management, Technology
Schools: Sloan '16 (D)
GMAT 1: 760 Q49 V44
GPA: 3.76
Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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New post 22 Nov 2011, 06:18
(3C2*10C3)/13C5 = 40/143

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Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls, [#permalink]

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Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls,   [#permalink] 28 Aug 2017, 21:25
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