You are not logged in.
Pages: 1
the probablity of getting a head on a single toss of a coin is p.consider that A starts and continues to flip the coin until a tail shows up, at which point B starts flipping. then B similarly continues until A takes over and so on. let P(n,m) be the probablity that A accumulates a total of n heads before B accumulates a total of m heads.
prove that:-
P(n,m)= pP(n-1,m)+(1-p)(1-P(m,n))
Offline
you can do a first step analysis for this question.
P(A gets n heads before B gets m heads)
= P(A gets n before B gets m | first toss is head)P(first toss is head) + P(A gets n before B gets m | first toss is tail)P(first toss is tail)
Cheers
Last edited by gckc123 (2009-11-14 11:02:00)
Maths is fun!
Offline
plz give the details ...
not able to understand
by d way
thanxxxx
but nothin to cheer for me
Offline
I used law of total probability and conditioned on the first toss.
Try to think about what the following probabilities are
P(A gets n heads before B gets m heads | first toss is head)
P(first toss is head)
Last edited by gckc123 (2009-11-14 16:54:51)
Maths is fun!
Offline
thanx dude........
at last i reached the solution
Offline
Pages: 1