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21 Nov 2011, 11:27
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:19) correct
26%
(02:20)
wrong
based on 135
sessions
History
Date
Time
Result
Not Attempted Yet
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?
A. 20/143
B. 28/143
C. 40/143
D. 60/143
E. 80/143
A. 20/143
B. 28/143
C. 40/143
D. 60/143
E. 80/143
Answer I got:
\(\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%\)
Is there another approach?
\(\frac{C^3_2*C^{10}_3}{C^{15}_5}=28%\)
Is there another approach?
arjunbt
Joined: 13 Jul 2011
Last visit: 27 Jul 2014
Posts: 68
Given Kudos: 42
Concentration: Operations, Strategy
Schools: Krannert '15 (WL) W.P. Carey (A)
GMAT 1: 680 Q46 V37
WE:Engineering (Telecommunications)
22 Nov 2011, 04:13
Kudos
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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
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
21 Nov 2011, 23:48
Kudos
Bookmarks
M3tm4n
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
