Question

Two persons $A$ and $B$ throw a coin alternatively till one of them gets head and wins the game. Find the respective probabilities of winning the game?

• A. $\frac{2}{3}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. $\frac{4}{7}$

Answer 1

The correct option is A $\frac{2}{3}$

Probabilities depends on who starts the game.

Let A starts the game

$P\left(Awinthe1stgame\right)$ $\frac{1}{2}$

Suppose $A$ does not win $1st$ game. For $A$ to participate in the 3rd game, a must lose 1st game and $B$ must be lose 2nd game.

Hence, this is conditional probability.

$\begin{array}{c}P\left(Awin3rdgamegivenAlsot1stgameandBlost2ndgame\right)\\ =P\left(Awin3rdgame\right)\times P\left(Alost1stgame\right)\times P\left(Blost2ndgame\right)\\ =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}={\left(\frac{1}{2}\right)}^3\\ Similarly,\\ P\left(Awin5thgamegivenAlost1stgame,Blost2ndgame,Alost3rdgameand4thgame\right)\\ =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}={\left(\frac{1}{2}\right)}^5\\ ThisprocesscontinuestillAorBwinthegame\\ Therefore\\ P\left(Awinsthegame\right)=\frac{1}{2}+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^5+.........\\ Thisisageometricalprogessionwithfirstterm.a=\frac{1}{2}andcommonratio,r={\left(\frac{1}{2}\right)}^2=\frac{1}{4}\\ So\\ P\left(Awinsthegame\right)=\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}\\ Hence,\\ obabilityofwhostartsthegamewins=\frac{2}{3}\\ obabilityoftheotherplayerwins=\frac{1}{3}\\ Hence,optionAisthecorrectanswer.\end{array}$