Re: DSE 2012- Question 42-44, 48 and 55-56.

Posted by wolverine on Apr 24, 2013; 10:13am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2012-Question-42-44-48-and-55-56-tp7580163p7580165.html

q42)  P(H)=P(T)=1/2

so prob of getting head in odd numbered toss wud be

p(H)+p(TTH)+p(TTTTH)+p(TTTTTTH)...so on

= 1/2 +1/2*1/2*1/2 + 1/2*1/2*1/2*1/2*1/2..so on...

 its a infinite gp with a=1/2 and r=(1/2)^2

sum of infinte  gp= a/(1-r)= (1/2)/(1-(1/2)^2)= 2/3