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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 21:47:42 IST
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a card from a pack of 52 cards is lost. from the remaining cards 2 cards are drawn & are found to be both diamond.find probability of lost card being diamond.
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18 Mar 2008 21:52:05 IST
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edited:see below
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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18 Mar 2008 22:03:05 IST
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but the ans is 11/50
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18 Mar 2008 22:03:53 IST
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very sorry...i picked only one card...
will do it...take a couple of minz
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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18 Mar 2008 22:05:26 IST
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Case I: lost card is a diamond
this happens in 13/52=1/4 ways
when this happens, chance of picking 2 diamonds is
12/51*11/50
so total=1/4*12/51*11/50=12*11/202*50
case II: non diamond is lost
3/4*13/51*12/50
=39*12/202*50
by Bayes theorem required
P=12*11/12*11+39*12
=11/11+39
=11/50
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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18 Mar 2008 22:09:54 IST
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E1 = lost card in a diamond
E2 = lost card nt a diamond
P(E1) = 13/52
P(E2) = 39/52
E = 2 cards drawn frm the remainin pack r diamonds
P(E/E1) = 12C2/51C2 = 12x11/51x50
P(E/E2) = 13C2/51C2 = 13x12/51x50
required propability
-----> P(E1/E) = P(E/E1) P(E1) / { P(E/E1) P(E1) + P(E/E2) P(E2)}
= 11/50
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"your future depends on what u do in present"
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18 Mar 2008 22:10:23 IST
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sry ... dint c akhil's answer.!
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