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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jul 2008 22:24:34 IST
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a die is rolled 3 times. the prob. of geting a no. larger than the previous no. each time is......
a) 5/72
b)5/54 (-> ans.)
c)13/216
d)1/18
somebody plz giv me the soln., i mean method.........
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26 Jul 2008 23:17:08 IST
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no on first roll second roll third roll total cases
6 5 1,2,3,4 4
6 4 1,2,3 3
6 3 1,2 2
6 2 1 1
5 4 1,2,3 3
5 3 1,2 2
5 2 1 1
4 3 1,2 2
4 2 1 1
3 2 1 1
thus total cases = 20
thus prob = 20/216
= 5/54
thats the answer.........
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nobody is perfect......i m nobody.............. |
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26 Jul 2008 23:31:59 IST
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3 nos can b chosen in 6C3 ways..
now they can b arranged in only one way side by side such that they are in ascending order.
so total ways such that each no is larger than the previous one =6C3..
now the total no of ways possible=6*6*6 ways
therefiore req prob = 6C3/6*6*6 = 20/ 6*6*6 = 5/ 54 .......
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B DESIGN
IIT GUWAHATI. |
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26 Jul 2008 23:42:22 IST
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Good work ankur, but great work by little genius. Superb
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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27 Jul 2008 18:30:21 IST
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HEY LITTLE GENIUS............ Plz explain me a little more elaborately, y u have taken dat 6C3.........
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30 Jul 2008 15:50:22 IST
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see to make all possible arr. of ascending ordered nos ,
we can do it by taking any possible combination amd then arranging it in ascending order.
first of all we find all the poss ways to choose 3 nos.
next we arrange tyhem in ascending order
since there is only one way to arrange any particular set of 3 nos so the total wways possible=total ways to choose 3 diff no sets=6C3
so prob=6C3/6*6*6
:)
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B DESIGN
IIT GUWAHATI. |
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31 Jul 2008 12:01:11 IST
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thanx genius.......... i finally got it....
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