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Please I need some intelligence,Patience...To help me in solving this question.Ty.
 one year ago
 one year ago
Please I need some intelligence,Patience...To help me in solving this question.Ty.
 one year ago
 one year ago

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EyadBest ResponseYou've already chosen the best response.1
An ant stands in the middle of a circle (3 metres in diameter) and walks in a straight line at a random angle from 0 to 360 degrees. Problem is, it can only walk one metre before it needs a break. The ant has the memory of a fish and forgets what direction it has just walked in.. Anyway, after the break, it gets all dizzy and thus chooses another random direction from 0 to 360 in an attempt to escape the circle again. As you can well imagine, it could escape the circle after just 2 walks (just one break needed). Or... it could take 20,000 walks (19,999 breaks needed)!! There might even be the very slim possibility it might take 20,000^20,000 walks.... What is the average amount of walks required for the ant to escape the circle?
 one year ago

EyadBest ResponseYou've already chosen the best response.1
@satellite73 ,@Calcmathlete
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
there is no snap solution to this problem i think try googling it and you will come up with many many posts, since apparently this is a common problem no good solutions though
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
am I on the completely wrong track to say that since he has a remaining 0.5 to go to escape the circle that \(\cos\theta\ge\frac12\implies\frac\pi6\le\theta\le\frac\pi6\)
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
not completely wrong....ur on the right track...:)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I mean \[\frac\pi3\le\theta\le\frac\pi3\]
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
yes...now u just compute the average
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
u do realize the ant neednt exactly go by ur condition!!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
so that is 1/3 of a circle... so there is a 1/3 chance he gets out on move 2 where do I include the average, I still don't quite get it :/
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
true...good enough. avearge is when u consider a particular no. of test cases. so if u consider 100, there is 33% chance of getting out on the 2nd move. now all u have to do is calculate the remaining chances for the other angles he might take....now although this is quite random, the probability for each angle range decreases n goes like a geometric series
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
yeah that's where the problem for me is... a geometric series is a good idea if we can set one up, but since the exact angle that he returns on is variable that won't be easy
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
like he could godw:1341369002782:dw(move 1)
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
yup...its a tedious problem...but i think a general solution is difficult but not impossible to obtain.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
dw:1341369028085:dwthat leaves him with a 0% chance of leaving the circle in the next move, ordw:1341369061211:dwin which case he has some other odds of leaving the circle in the next move
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
@Zarkon probability problem, interested?
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
for all u know, there is even the possiblity like the question itself that the ant might not come out at all if he keeps taking angles that put him back inside.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
that was suggested by @Outkast3r09 on an earlier post, but @Eyad says there is a definite answer, so...
 one year ago

amrit110Best ResponseYou've already chosen the best response.0
http://math.stackexchange.com/questions/133851/antinacircle. this page has a good solution.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
well that still is not as elegant as I was hoping where's the actual answer now...?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
11.4 is the "expected value" so I guess that's the answer
 one year ago

EyadBest ResponseYou've already chosen the best response.1
Ok Then Ty everyone :) Ty @TuringTest ,@amrit110 :ty for the site its useful :)
 one year ago
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