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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Dec 2007 15:10:58 IST
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A bag contains 20 coins. If the prob. that bag contains 4 biased coins is 1/3 and that of
exactly 5 biased coin is 2/3 then the prob. that all the biased coin are sorted out from the
bag in exactly 10 draws is???
rates assured
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lol... |
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5 Dec 2007 19:50:21 IST
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Koi to kar de
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lol... |
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7 Dec 2007 21:36:01 IST
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admin pls note change this to maths section
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" You get What u deserve and not what you desire"
R.darshan kumar |
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7 Dec 2007 22:11:36 IST
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is it 7/255.....if it is ,then i'll explain
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"Nenenthedhavano naake teleedu"
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7 Dec 2007 22:15:54 IST
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im not clear wid the question....while solving how many biased coins do we have to consider 4 or 5?? or do u mean that their r either 4 or 5 coins in the bag???can u plz specify
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7 Dec 2007 23:21:37 IST
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see there cant be 4 and 5 at a time and also notice that the probabilities given for them are 1/3 and 2/3 (i mean they add up to 1)means they are exhaustive and exclusive events ........this qn. is based on addition theorem
......hence the answer will be P(there are exactly 4biased coins)P(test ends at 10th draw given there are 4 biased coins)+P(there are exactly 5 biased coins)P(test ends at 10th draw given there are 5 biased coins).......understood? if not nudge me .....i'll make it clear
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"Nenenthedhavano naake teleedu"
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8 Dec 2007 00:07:48 IST
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Probability that all biased coins would be sorted out at the 10th position when 4 biased coins are there = P4 Probability that all biased coins would be sorted out at the 10th position when 5 biased coins are there = P5 Total Probability = P4+P5 P4 = 1/3 * [(4C3*16C6)/(20C9)] * (1/11) i.e the probabilty that out of the 4 biased coins, exactly 3 come before 10th position and out of 16 unbiased coins exactly 6 come before 10th pos and the last biased coin comes at 10th pos. Similarly P5 = 2/3 *[(5C4*15C5)/20C9] * (1/11) Total prob = P4+P5 which comes out to be 53/3876.......(ans)
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There is no better feeling in this world than being a winner! |
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8 Dec 2007 13:06:20 IST
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let E4 be the event that there are exactly 4 biased coins
let E5 be the event that there are exactly 5 biased coins
let E be the event that process ends at 10th draw
now P(E)=P(E4)P(E/E4) + P(E5)P(E/E5)
P(E/E4)=[(4C316C6)/(20C9)](1/11)
P(E/E5)=[(5C415C5)/20C9](1/11)
P(E4)=1/3
P(E5)=2/3
=>P(E)=(1/3)[(4C316C6)/(20C9)](1/11) + (2/3)[(5C415C5)/20C9](1/11)
=>P(E)=217/19380
hope u understood now.......and in my previous answer(7/255) there is a calculation mistake...now this is correct
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"Nenenthedhavano naake teleedu"
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