\(A_{i};i=1,2,3..n\) are \(n\) persons which play a game of tossing a coin. The game starts when \(A_{1}\) tosses the coin. If he gets a tail, the next person i.e. \(A_{2}\) gets a chance to toss the coin and so on. Whosoever gets a head first wins the game. If no one wins, the game is played again. Find the probability of event in which the \({r}^{th}\) person i.e. \(A_{r}\) wins the game.

## Comments

Sort by:

TopNewestThe probability that the game is played again is \(\dfrac{1}{2^{n}}\).The probability that the r th person wins in the x th round of the game is \(\dfrac{1}{2^{xn+r}}\).Hence the answer,according to me should be,\(\sum_{x=1}^{\infty}\dfrac{1}{2^{xn+r}}\). This is a simple G.P. What do you think? – Adarsh Kumar · 1 year, 5 months ago

Log in to reply

– Rohit Ner · 1 year, 5 months ago

oh yeah! i too got the same answer! cheers!Log in to reply

– Adarsh Kumar · 1 year, 5 months ago

Cheers!From where did you get this question?Log in to reply

– Rohit Ner · 1 year, 5 months ago

Actually the question struck to my mind when my sir was teaching us conditional probability.its a generalized version I managed to frame. :PLog in to reply

– Adarsh Kumar · 1 year, 5 months ago

OHk!Log in to reply

Log in to reply

– Rohit Ner · 1 year, 5 months ago

I am not aware of the answer. Could you please post you work?Log in to reply

Log in to reply

– Rohit Ner · 1 year, 5 months ago

Oh the problem is not supposed to be like that..wait let me specify it.sorry for it.Log in to reply

– Adarsh Kumar · 1 year, 5 months ago

No problemo!Log in to reply