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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Jan 2008 23:50:30 IST
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JEE 2004 A BAG CONTAINS 12 RED BALLS AND 6 WHITE BALLS. 6 BALLS ARE DRAWN ONE BY ONE WITHOUT REPLACEMENT OF WHICH ATLEAST 4 BALLS ARE WHITE. FIND THE PROBABILITY THAT IN THE NEXT TWO DRAWS EXACTLY 1 WHITE BALL IS DRAWN?  LEAVE THE ANSWER IN TERMS OF nCr
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FAILURE, THE FIRST STEP TO SUCCESS |
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17 Jan 2008 13:06:35 IST
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After the first 6 draw , we have the following mutually exclusive & exhaustive possibility 1. R : 10 W :2 2 . R : 11 , W : 1 3. R :12 , W : 0 For the first case the probability ,that after the subsequent 2 draws , 1 white ball is drawn is =P ( W R ) + P ( RW ) ( the ordered pair denotes their relative order of occurance ) = 2/ 12 * 10/11 + 10/12 * 2/11 = 2 * 2 *10/( 12 *11 ) similarly for case 2 we have probability = 2 * 1 * 11 / ( 12 * 11 ) for the last case the probability is zero So the total probability = 2 * ( 20 + 11 ) / ( 12 * 11 ) = 62 /132 = 31 / 66 ( Ans )
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18 Jan 2008 15:27:30 IST
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THE PROBLEM IS THAT THE ANSWER GIVEN IN THE BOOK IS IN TERMS OF nCr AND IT IS QUITE A BIG EXPRESSION
ANYWAYS I DID NUMERICALLY SIMPLIFY THE ANSWER AND IT IS ALMOST HALF THE ANSWER YOU GOT
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FAILURE, THE FIRST STEP TO SUCCESS |
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18 Jan 2008 16:57:29 IST
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refer fiitjee.com iit-jee 2004 mains soln. it's explained nicely ....
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"Imagination is more important than knowledge."
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18 Jan 2008 17:45:13 IST
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Let P(A) be the prob that at least one white ball be
drawn and P(B) be the probability that exactly 1 white
ball is drawn.
P(B/A) = [I=1] [3] P(Ai) P(B/Ai) / [I=1] [3] P(Ai)
= [(12C2* 6C4/18C6) * (10C1*2C1/12C2) + (12C1*
6C5/18C6 ) *(11C1*1C1/12C2)] / [ (12C2*6C4/18C6 +
12C1*6C5/18C6 + 12C0 * 6C6/18C6)]
Now simplify to obtain the answer
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All that is gold does not glitter
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The old that is strong does not wither,
Deep roots are not reached by frost.
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The crown less again shall be king. |
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18 Jan 2008 21:15:00 IST
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Can u pleaze point out the bug in my solution.
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18 Jan 2008 22:41:47 IST
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@feynmann a major bug in ur soln is that u havn't found out the how the given cases can arise. i.e how 4 , or 5 , or 6 white balls are selected in 6 draws.
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18 Jan 2008 23:11:22 IST
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O. K. got it . Then attach the probabilty factor to each three cases .
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