A bag contains a white and b black balls. Two players A and B alternately 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. If the probability of A winning the game is three times that of B, then the ratio of $a:b$ is ;
$\left( 1 \right)1:1$
$\left( 2 \right)1:2$
$\left( 3 \right)2:1$
$\left( 4 \right)$ None of these

Answer
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Hint: This question deals with the basic concepts of probability. Probability is always calculated in cases whenever we are not certain about the outcomes of an event. The probability of an event can be calculated as : $P\left( {{\text{Event}}} \right) = \dfrac{{{\text{Favourable outcomes}}}}{{{\text{Total outcomes}}}}$ . For example: If an unbiased coin is tossed, then the probability of getting a head is ; $P\left( {{\text{head}}} \right) = \dfrac{1}{2}{\text{ means 50}}\% $ ( here total outcomes are $2$ i.e. either heads or tail ) . Probability and information are inversely proportional to each other means if the probability of an event is more then the information related to that event will be less. For example: a dog bites a man , this event is common in occurrence and hence the information will be less or limited but if the reverse event happens i.e. a man bites a dog then the information will be high in that case as the probability of such event is very less.


Complete step by step answer:
Given information: $a = $ Number of white balls ; $b = $ Number of black balls .
Total number of balls are : $a + b$
There are two players A and B and the probability of winning A is three times that of B means, ${\text{A = 3B}}$
$\left( 1 \right)$ The probability of drawing a white ball from the bag is ;
$ \Rightarrow {\text{P}}\left( {{\text{white}}} \right) = {\text{ }}\dfrac{a}{{a + b}}{\text{ }}......\left( 1 \right)$
$\left( 2 \right)$ The probability of drawing a black ball from the bag is ;
$ \Rightarrow {\text{P}}\left( {{\text{black}}} \right) = {\text{ }}\dfrac{b}{{a + b}}{\text{ }}......\left( 2 \right)$
According to the given question, player A starts the game and players A and B are drawing the ball from the bag alternately and if a player draws a white ball from the bag then that player wins and the game stops;
Let us discuss the probabilities of player A winning the game ;
$\left( {\text{i}} \right){\text{P}}\left( {\text{w}} \right) = $ Player A starts the game and the first ball drawn is white, A wins and game stops.
$\left( {{\text{ii}}} \right){\text{P}}\left( {{\text{bbw}}} \right) = $ Player A draws black ball then player B draws black ball again and then A draws the white ball, A wins the game and game is stopped.
Similarly, this concept can go on like this.
Hence, the probability that player A wins the game is ;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right){\text{ = P}}\left( {{\text{w, bbw, bbbw, bbbbw, }}...............} \right)$
The above concept can be written as;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right){\text{ = P}}\left( {\text{w}} \right) + {\text{P}}\left( {{\text{bbw}}} \right) + {\text{P}}\left( {{\text{bbbw}}} \right) + {\text{P}}\left( {{\text{bbbbw}}} \right) + ........$
One important point to note in this question is that each time a ball is drawn by a player it is replaced, which indicates both the events are independent events meaning that the probability of occurrence of one event does not effect the probability of occurrence of another event and we know that if A and B are two independent events, then the formula ;
$ \Rightarrow {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)$
By the above formula, we can expand our equation as;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right){\text{ = P}}\left( {\text{w}} \right) + {\text{P}}\left( {{\text{b}} \cap {\text{b}} \cap {\text{w}}} \right) + {\text{P}}\left( {{\text{b}} \cap {\text{b}} \cap {\text{b}} \cap {\text{w}}} \right) + {\text{P}}\left( {{\text{b}} \cap {\text{b}} \cap {\text{b}} \cap {\text{b}} \cap {\text{w}}} \right) + ........$
The above infinite geometric progression can also be written as;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right){\text{ = P}}\left( {\text{w}} \right) + \left[ {{\text{P}}\left( {\text{b}} \right) \times {\text{P}}\left( {\text{b}} \right) \times {\text{P}}\left( {\text{w}} \right)} \right] + \left[ {{\text{P}}\left( {\text{b}} \right) \times {\text{P}}\left( {\text{b}} \right) \times {\text{P}}\left( {\text{b}} \right) \times {\text{P}}\left( {\text{w}} \right)} \right] + ........$
This is an infinite geometric progression with, First term $ = {\text{P}}\left( {\text{w}} \right)$ and common ratio $ = {\text{P}}{\left( {\text{b}} \right)^2}$ ;
By the formula of an infinite geometric series, the sum of infinite terms is given by ;
$ \Rightarrow {S_\infty } = \dfrac{a}{{a - r}}$
Applying the formula to the above equation, we get;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = \dfrac{{{\text{P}}\left( {\text{w}} \right)}}{{1 - {\text{P}}{{\left( {\text{b}} \right)}^2}}}$
Putting the values of ${\text{P}}\left( {\text{w}} \right)$ and ${\text{P}}\left( {\text{b}} \right)$ from equation $\left( 1 \right){\text{ and }}\left( 2 \right)$ , we get ;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = \dfrac{{\dfrac{a}{{a + b}}}}{{1 - {{\left( {\dfrac{b}{{a + b}}} \right)}^2}}}$
Simplifying the above expression ;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = \dfrac{{\dfrac{a}{{a + b}}}}{{\dfrac{{{{\left( {a + b} \right)}^2} - {b^2}}}{{{{\left( {a + b} \right)}^2}}}}}$
By the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , simplifying the above expression ;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = a \times \dfrac{{a + b}}{{{a^2} + {b^2} + 2ab - {b^2}}}$
On further simplification;
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = \dfrac{{a\left( {a + b} \right)}}{{a\left( {a + 2b} \right)}}$
$ \Rightarrow {\text{P}}\left( {{\text{A wins the game}}} \right) = \dfrac{{a + b}}{{a + 2b}}{\text{ }}......\left( 3 \right)$
We can calculate the probability of player B winning the game as ;
$ \Rightarrow {\text{P}}\left( {{\text{B wins the game}}} \right) = 1 - {\text{P}}\left( {{\text{A wins the game}}} \right)$
$ \Rightarrow {\text{P}}\left( {{\text{B wins the game}}} \right) = 1 - \dfrac{{a + b}}{{a + 2b}}$
Simplifying the above expression;
$ \Rightarrow {\text{P}}\left( {{\text{B wins the game}}} \right) = \dfrac{b}{{a + 2b}}{\text{ }}......\left( 4 \right)$
Now, comparing equation $\left( 3 \right){\text{ and }}\left( 4 \right)$ ; and also we know that Probability of winning player A is 3 times that of player B i.e. A = 3B ;
$ \Rightarrow \dfrac{{a + b}}{{a + 2b}} = 3\left( {\dfrac{b}{{a + 2b}}} \right)$
$ \Rightarrow a + b = 3b$
$\therefore a = 2b$
Therefore, the ratio of $a:b = 1:2$

So, the correct answer is “Option 2”.

Note:
$\left( 1 \right)$ Independent events: Two events are said to be independent if the outcome of an event does not affect the outcome of another event means they are mutually exclusive and are related by formula: $P\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)$ . $\left( 2 \right)$ Dependent events: Two events are said to be dependent if the outcome of one event affects the outcome of another event and they are related by the formula : $P\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\left. {B{\text{ }}} \right|{\text{A}}} \right)$ where ${\text{P}}\left( {\left. {B{\text{ }}} \right|{\text{A}}} \right)$ represents the probability of occurrence of event B given that event A has already occurred.