Question

A bag contains ‘a’ white balls and ‘b’ black balls. Two players, A and B alternatively draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game and if the probability of winning the game is three times that of B, then find the ratio of $a:b$?
(a) $1:1$,
(b) $1:2$,
(c) $2:1$,
(d) None of these.

Answer 1

We start solving the problem by finding the probability of drawing a white and black ball from the bag. We then find the probability that A wins the game by using the fact A wins the game if he wins in 1st game or 3rd game or 5th game (if B doesn’t win). We then find the probability that B wins the game if he wins in the 2nd game or 4th game or 6th game (if A doesn’t win). We then take the ratio of these for using the data that the probability of winning the game is three times that of B and make necessary calculations to get the required ratio of a and b.

Complete step-by-step answer:
According to the problem, we have a bag with ‘a’ white balls and ‘b’ black balls. Two players, A and B alternatively draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game and A begins the game. We need to find the value of $a:b$ if the probability of winning the game is three times that of B.
We have a total of $\left( a+b \right)$ balls in the bag. Let us assume the probability of drawing the white and black balls is $P\left( a \right)$ and $P\left( b \right)$.
So, we have $P\left( a \right)=\dfrac{\text{No}\text{. of white balls in bag}}{\text{Total no}\text{. of balls in bag}}$.
$\Rightarrow P\left( a \right)=\dfrac{\text{a}}{\text{a+b}}$ ---(1).
Now, we have $P\left( b \right)=\dfrac{\text{No}\text{. of black balls in bag}}{\text{Total no}\text{. of balls in bag}}$.
$\Rightarrow P\left( b \right)=\dfrac{\text{b}}{\text{a+b}}$ ---(2).
Let us assume the probability of A winning the game is $P\left( A \right)$ and $P\left( B \right)$.
According to the problem, A begins the game. Now, we find the probability that A wins the game.
A wins the game if he wins in the 1st game or 3rd game or 5th game (if B doesn’t win). We know that drawing balls is an independent event.
So, we have $P\left( A \right)=P\left( a \right)+P\left( b \right)P\left( b \right)P\left( a \right)+P\left( b \right)P\left( b \right)P\left( b \right)P\left( b \right)P\left( a \right)+............\infty$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a}{a+b} \right)+\left( \left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{a}{a+b} \right) \right)+\left( \left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{a}{a+b} \right) \right)+............\infty$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a}{a+b} \right)\left( 1+{{\left( \dfrac{b}{a+b} \right)}^{2}}+{{\left( \dfrac{b}{a+b} \right)}^{4}}+............\infty \right)$.
We know that the probability lies between 0 and 1. We know that the sum of geometric series up to infinity for common ratio less than 1 is $\dfrac{a}{1-r}$, where a is first term and r is common ratio $\left( \left| r \right|<1 \right)$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a}{a+b} \right)\left( \dfrac{1}{1-{{\left( \dfrac{b}{a+b} \right)}^{2}}} \right)$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a}{a+b} \right)\left( \dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a+b \right)}^{2}}-{{\left( b \right)}^{2}}} \right)$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a\left( a+b \right)}{\left( a+2b \right)\left( a \right)} \right)$.
$\Rightarrow P\left( A \right)=\left( \dfrac{a+b}{a+2b} \right)$ ---(3).
Now, we find the probability that B wins the game.
B wins the game if he wins in the 2nd game or 4th game or 6th game (if A doesn’t win). We know that drawing balls is an independent event.
So, we have $P\left( B \right)=P\left( b \right)P\left( a \right)+P\left( b \right)P\left( b \right)P\left( b \right)P\left( a \right)+P\left( b \right)P\left( b \right)P\left( b \right)P\left( b \right)P\left( b \right)P\left( a \right)+............\infty$.
$\Rightarrow P\left( B \right)=\left( \left( \dfrac{b}{a+b} \right)\left( \dfrac{a}{a+b} \right) \right)+\left( \left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{a}{a+b} \right) \right)+\left( \left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{b}{a+b} \right)\left( \dfrac{a}{a+b} \right) \right)+...\infty$.
$\Rightarrow P\left( B \right)=\left( \dfrac{ab}{{{\left( a+b \right)}^{2}}} \right)\left( 1+{{\left( \dfrac{b}{a+b} \right)}^{2}}+{{\left( \dfrac{b}{a+b} \right)}^{4}}+............\infty \right)$.
We know that the probability lies between 0 and 1. We know that the sum of geometric series up to infinity for common ratio less than 1 is $\dfrac{a}{1-r}$, where a is first term and r is common ratio $\left( \left| r \right|<1 \right)$.
$\Rightarrow P\left( B \right)=\left( \dfrac{ab}{{{\left( a+b \right)}^{2}}} \right)\left( \dfrac{1}{1-{{\left( \dfrac{b}{a+b} \right)}^{2}}} \right)$.
$\Rightarrow P\left( B \right)=\left( \dfrac{ab}{{{\left( a+b \right)}^{2}}} \right)\left( \dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a+b \right)}^{2}}-{{\left( b \right)}^{2}}} \right)$.
$\Rightarrow P\left( B \right)=\left( \dfrac{ab}{\left( a+2b \right)\left( a \right)} \right)$.
$\Rightarrow P\left( B \right)=\left( \dfrac{b}{a+2b} \right)$ ---(4).
According to the problem, we have $P\left( A \right):P\left( B \right)=3:1$.
$\Rightarrow \dfrac{\left( \dfrac{a+b}{a+2b} \right)}{\left( \dfrac{b}{a+2b} \right)}=\dfrac{3}{1}$.
$\Rightarrow \dfrac{\left( a+b \right)}{\left( b \right)}=\dfrac{3}{1}$.
$\Rightarrow a+b=3b$.
$\Rightarrow a=2b$.
$\Rightarrow \dfrac{a}{b}=\dfrac{2}{1}$.
∴ The ratio of a and b is $2:1$.

So, the correct answer is “Option c”.

Note: We should not make mistakes while calculating the probabilities of winning of A and B. We should know that in these cases the probability can be equal to zero winning of B if A wins in the first game. We can also find the approximation of the total number of balls in the bag using the ratio that we have just obtained. Similarly, we can also expect problems which contain more than 2 players.