Question

\[A\] and \[B\] toss a coin alternatively. The first to show a head being the winner. If \[A\] starts a game, then what is the chance of his winning?
A. \[\dfrac{5}{8}\]
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{3}\]
D. \[\dfrac{2}{3}\]

Answer 1

Hint First, calculate the probability of getting a head or a tail after tossing a coin. Then calculate the winning probability of the player \[A\] using the given condition.

Formula used:
Sum of infinite terms in geometric progression: \[{S_\infty } = \dfrac{a}{{1 - r}}\] , \[ - 1 < r < 1\]. Where \[a\] is the first term and \[r\] is the common ratio.
Complement rule: \[P\left( A \right) + P\left( {A'} \right) = 1\] , where \[P\left( {A'} \right)\] is the complement of \[P\left( A \right)\]

Complete step by step solution:
Given:
Two players \[A\] and \[B\] toss a coin alternatively and \[A\] starts a game.
Let consider,
\[H :\] Event of getting a head
\[T:\] Event of getting a tail

The probability of getting a head is, \[P\left( H \right) = \dfrac{1}{2}\]
The probability of not getting a head is, \[P\left( T \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}\]

Since \[A\] begins the game. Player \[A\] wins if he gets a head in the \[{1^{st}}\] throw or in the \[{3^{rd}}\] throw or in the \[{5^{th}}\] throw and so on.
So, the probability that player \[A\] wins the game is
\[P\left( A \right) = P\left( \text{H or TTH or TTTTH or....} \right)\]
\[ \Rightarrow \]\[P\left( A \right) = P\left( H \right) + P\left( T \right)P\left( T \right)P\left( H \right) + P\left( T \right)P\left( T \right)P\left( T \right)P\left( T \right)P\left( H \right) + ....\]
\[ \Rightarrow \]\[P\left( A \right) = P\left( H \right) + P\left( H \right){\left( {P\left( T \right)} \right)^2} + P\left( H \right){\left( {P\left( T \right)} \right)^4} + ....\]

The terms of the above addition are in geometric progression with the first term \[a = P\left( H \right)\] common ratio \[r = {\left( {P\left( T \right)} \right)^2}\].
The value of \[P\left( T \right)\] is less than 1. So, the value of \[r\] is also less than 1.
Apply the formula of the sum of infinite terms in geometric progression.
\[P\left( A \right) = \dfrac{{P\left( H \right)}}{{1 - {{\left( {P\left( T \right)} \right)}^2}}}\]
Substitute the values of \[P\left( H \right)\] and \[P\left( T \right)\] in the above equation.
\[P\left( A \right) = \dfrac{{\dfrac{1}{2}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}}\]
Simplify the above equation.
\[P\left( A \right) = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{4}}}\]
\[ \Rightarrow \]\[P\left( A \right) = \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{3}{4}} \right)}}\]
\[ \Rightarrow \]\[P\left( A \right) = \dfrac{1}{{\left( {\dfrac{3}{2}} \right)}}\]
\[ \Rightarrow \]\[P\left( A \right) = \dfrac{2}{3}\]
Hence the correct option is D.

Note: Students often get confused about the formula of the sum of infinite terms in the geometric progression.
The formula of the sum of infinite terms in the GP, \[{S_\infty } = \dfrac{a}{{1 - r}}\] is used when \[\left| r \right| < 1\].
If the common ratio \[\left| r \right| \ge 1\], then the sum is infinite and the series is divergent.