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   Author  Topic: i need help  (Read 881 times)
boom
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i need help  
« on: Apr 16th, 2004, 8:20pm »
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Thanks for problem one
 
problem 2
given a discrete source of  
 [smiley=lceil.gif] X   [smiley=rceil.gif] [smiley=eq.gif]   [smiley=lceil.gif] a1=0  a2=1  a3=2  a4=0  [smiley=rceil.gif]
 [smiley=lfloor.gif]p(x) [smiley=rfloor.gif]      [smiley=lfloor.gif] 3/8    1/4     1/4    3/8    [smiley=rfloor.gif]
 
the output message is (202120130213001203210110321010021032011223210 .find
1.the self-information in this message
2.the information contained in per outcome of this message
 
problem 3
A dishonest gambler has a loaded dice which turns up the number 1 with probability 2/3 and the numbers 2 to 6 with probability 1=1/15 each.unfortunately,he has left his loaded dice in a box with two honest dices and not tell them apart.
 
(a) He picks one dice(at random) from the box,rolls it once and the number 1 appears.conditional on this result,what is the probability that he picked up the loaded dice?
 
(b)He now rolls the dice once more and it comes up 1 again .what is  the probability after this scecond rolling that he has picked up the loaded die?
 
« Last Edit: Apr 17th, 2004, 9:59pm by boom » IP Logged
towr
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Re: i need help  
« Reply #1 on: Apr 17th, 2004, 6:53am »
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1.1 3/4 (the maximum would be if they are disjoint).
1.2 37/64 (all are independant, so 1-(1-1/4)3 ).
 
I'm not sure what problem 2 means..
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Icarus
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Re: i need help  
« Reply #2 on: Apr 17th, 2004, 10:00am »
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There seems to be some numbering confusion in your problems.
 
The answers towr gives are for part 2 of your first problem, not part 1.
 
(problem 1.1): if E1, E2, and E3 are just 3 names for the same event, then E0 is yet another name for the same event. Hence P(E0) = 1/4 as well.
 
(problem 1.2.1): 3/4
 
(problem 1.2.2): 37/64
 
And I also cannot make any sense of problem 2. Though this is to be expected since towr is far more knowledgable than I am in this area.
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