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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
Joined: 06 Nov 2011
Posts: 37
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
GPA: 3
Followers: 0

Kudos [?]: 7 [0], given: 50

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

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21 Nov 2011, 11:27
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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 1 sessions

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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?

[Reveal] Spoiler:
$$\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%$$

Is there another approach?
Manager
Joined: 31 May 2011
Posts: 88
Location: India
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 48 [0], given: 4

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

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21 Nov 2011, 12:14
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?

[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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6970
Location: Pune, India
Followers: 2027

Kudos [?]: 12753 [0], given: 221

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

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21 Nov 2011, 23:48
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?

[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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Manager
Joined: 13 Jul 2011
Posts: 151
Concentration: Operations, Strategy
GMAT 1: 680 Q46 V37
WE: Engineering (Telecommunications)
Followers: 1

Kudos [?]: 35 [1] , given: 42

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

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22 Nov 2011, 04:13
1
KUDOS
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
Manager
Joined: 29 Oct 2011
Posts: 184
Concentration: General Management, Technology
Schools: Sloan '16 (D)
GMAT 1: 760 Q49 V44
GPA: 3.76
Followers: 10

Kudos [?]: 130 [0], given: 19

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

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22 Nov 2011, 07:18
(3C2*10C3)/13C5 = 40/143
Re: There are 3 black balls and 10 white balls. If one is to pick 5 balls,   [#permalink] 22 Nov 2011, 07:18
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