Math Wiz: probability!

[Bluga]

*I think the hard part is to understand the question

-A robot is controlled by an instruction tape, on which a string of bits of length n is recorded.
-In one unit time, the robot reads a bit from the tape (after which the tape is advanced to the next bit position).
-If the bit is a one, the robot takes one step towards the door, and if the bit is a zero, the robot remains where it is.
-This process is then repeated during the next unit of time until all bits have been read. Assume that the tape contains a complete random string of bits, each bit being one or zero with equal probability.

a)How many possible instruction tapes are there of length n?

b)how many different instruction tapes of length n result in a movement of k steps towards the door?

c)After the robot has read in n bits, for each k in the range 0<=k<=n, find the probability that it has moved k stpes towards the door.

d)Supose n=12, and that the robot will pass through the door if it takes 8 or more steps. What's the probability that the robot will pass through the door?

 

[Sigurd]

Do your own homework man!
Good ol binomial distribution. Too simple to waste time with

 

[scauffiel]

Wha??

...these aren't the droids you're looking for...

 

[Soccer55]

I'm not going to do it for you, but here's some help:
a) There are 2 possibilities for each bit (0 or 1) and there are n bits in each tape. How do you put those together to get the total number of tapes?
b)If you know how many total bits there are in a tape and how many steps you need, you can figure out how many of the tapes will have that many steps just by examining the possibilities for each case from k to n and summing them.
c)As Sigurd said, you will probably need a binomial random variable and find the distribution.
d)Same as c) except with numbers in place of k and n.

This should be right and get you going in the right direction.

-Tom

 

[Jzero]

I have the sudden uncontrollable urge to play roborally....

 

[Bluga]

<< a) There are 2 possibilities for each bit (0 or 1) and there are n bits in each tape. How do you put those together to get the total number of tapes?
b)If you know how many total bits there are in a tape and how many steps you need, you can figure out how many of the tapes will have that many steps just by examining the possibilities for each case from k to n and summing them.
c)As Sigurd said, you will probably need a binomial random variable and find the distribution.
d)Same as c) except with numbers in place of k and n. >>




a) 2^n ??
b)C(n,k) ??
c) a)/b) ??

what i'm not sure is isn't there only "ONE" tape? why is it asking "different instruction tapes"??

 

[Bluga]

bump

 

[crypticlogin]

<< what i'm not sure is isn't there only "ONE" tape? why is it asking "different instruction tapes"?? >>



I'm pretty sure they're asking how many variations of length-n tape can exist. That should be simple enough (and if not, forget the binomial distribution and bone up on the combinatorics, permutations, and the simpler prob. problems)... but this is like the de facto example problem for a binomial distribution.