Probability Challenge!

$A_{i};i=1,2,3..n$ 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.

#Combinatorics #Probability #ConditionalProbability #IITJEE

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

The 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 - 5 years, 8 months ago

oh yeah! i too got the same answer! cheers!

Rohit Ner - 5 years, 8 months ago

Cheers!From where did you get this question?

Adarsh Kumar - 5 years, 8 months ago

@Adarsh Kumar – Actually the question struck to my mind when my sir was teaching us conditional probability.its a generalized version I managed to frame. :P

Rohit Ner - 5 years, 8 months ago

@Rohit Ner – OHk!

Adarsh Kumar - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$