PROBABILITY THAT HEADS WILLE UP WHEN a FAIR COIN IS TOSSED?

Let 惟 = {w1,w2,w3...wn} , wi is an outcome in the sample space and


X be a discrete random variable over S with probability mass function


m(w):s--%26gt;[0,1]


then 危 (wi) | i=1..n =1. where m(w3)=...m(wn)=0.


The moment generating function of X is M_X(t)=EXP(e^(tX)), t E R


=危 (wi)


Let 胃 = { }. Then w1 /\ w2 = $theta;


Corillary: EXP(X) for a random variable X over a probabilitically symmetrical, binary and uniform S.


= Pr(X=wi) for each i.


=%26gt;危 e^(tw)*m(w) , w E Z= M_X(t)


where dM(t)/dt|t=0 =EXP(X). therefore


危 w*m(w) , w E Z = 1/2*1+0*1/2=1/2


Answer = 1/2





??????????????|||Erm... right. A coin has two sides. The probability that any one side will come up is 1/2 (the required side/the number of sides), so for n tosses, the probability that it will come up the same side every time is 1/(2^n).|||You are an idiot dude. you went thru all that and you kno there are 2 sides. That means you have a 50 percent chance of gettiing 1 side.