Re: Isi 2015 PEA Q 26.

Posted by Akash on Apr 19, 2016; 1:50pm
URL: http://discussion-forum.276.s1.nabble.com/Isi-2015-PEA-Q-26-tp7599922p7599952.html

Let n be money left with Priya at any given time.
And,  p(n) is her chance chances of winning the total amount a + b (all of it)

p(0)=0 [she has lost]
p(n) = p(n+1)/2 + p(n-1)/2 [at every trial, win or lose one rupee, both with the probability of 1/2]
2p(n) = p(n+1) + p(n-1)
p(n+1) - p(n) = p(n) - p(n-1)

By recursion,
p(n) = n p(1)

We know that p(a + b) = 1
p(1) = 1 / (a + b)
p(a)=a*p(1)=a/(a+b)