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        riddles >> general problem-solving / chatting / whatever >> i need help 
        
(Message started by: boom on Apr 16^th, 2004, 8:20pm)

Title: i need help
Post by boom on Apr 16^th, 2004, 8:20pm

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?

Title: Re: i need help
Post by towr on Apr 17^th, 2004, 6:53am

1.1 [hide]3/4 (the maximum would be if they are disjoint)[/hide].
1.2 [hide]37/64 (all are independant, so 1-(1-1/4)³ )[/hide].

I'm not sure what problem 2 means..

Title: Re: i need help
Post by Icarus on Apr 17^th, 2004, 10:00am

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.