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Hint: Try to find out probability in all those possible cases in which $A$ wins and then take the union of all these cases by simply adding these probabilities. To add the probabilities, use a formula for the sum of infinite G.P.

In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total

of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,

$P\left( 6 \right)=\dfrac{1}{6}$

Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not

getting $6$ on the dice.

$\Rightarrow P\left( X \right)=\dfrac{1}{6}$

And $P\left( Y \right)=\dfrac{5}{6}$

To find the probability that $A$ wins, let us make all the possible cases.

Case 1: $A$ wins at first throw of dice

$\begin{align}

& P\left( A-wins \right)=P\left( X \right) \\

& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\

\end{align}$

Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,

$A$ will throw the dice.

Case 2: $A$ wins at the third throw of dice

$\begin{align}

& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{

}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\

& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\

\end{align}$

Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.

Case 3: $A$ wins at the fifth throw

$\begin{align}

& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{

}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}

\right).P\left( six\text{ }by\text{ }A \right) \\

& \Rightarrow P\left( A\text{ }wins

\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\

& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\

\end{align}$

Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability

of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .

Since all the above listed cases are possible, the probability of $A$ winning is given by the union of

all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,

\[\begin{align}

& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(

\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}

\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........

\right)............\left( 1 \right) \\

\end{align}\]

The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}

\right)}^{2}}\] which is less than $1$, we can find it’s sum.

For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’

${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$

Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with

$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,

\[\begin{align}

& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)

\\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\

\end{align}\]

In probability, we have a property,

$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4

\right)$

Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,

$\begin{align}

& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\

& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\

\end{align}$

From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above

equation, we get,

$\begin{align}

& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\

& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\

\end{align}$

Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.

In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total

of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,

$P\left( 6 \right)=\dfrac{1}{6}$

Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not

getting $6$ on the dice.

$\Rightarrow P\left( X \right)=\dfrac{1}{6}$

And $P\left( Y \right)=\dfrac{5}{6}$

To find the probability that $A$ wins, let us make all the possible cases.

Case 1: $A$ wins at first throw of dice

$\begin{align}

& P\left( A-wins \right)=P\left( X \right) \\

& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\

\end{align}$

Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,

$A$ will throw the dice.

Case 2: $A$ wins at the third throw of dice

$\begin{align}

& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{

}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\

& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\

\end{align}$

Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.

Case 3: $A$ wins at the fifth throw

$\begin{align}

& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{

}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}

\right).P\left( six\text{ }by\text{ }A \right) \\

& \Rightarrow P\left( A\text{ }wins

\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\

& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\

\end{align}$

Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability

of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .

Since all the above listed cases are possible, the probability of $A$ winning is given by the union of

all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,

\[\begin{align}

& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(

\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}

\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........

\right)............\left( 1 \right) \\

\end{align}\]

The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}

\right)}^{2}}\] which is less than $1$, we can find it’s sum.

For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’

${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$

Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with

$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,

\[\begin{align}

& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)

\\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\

& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\

\end{align}\]

In probability, we have a property,

$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4

\right)$

Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,

$\begin{align}

& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\

& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\

\end{align}$

From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above

equation, we get,

$\begin{align}

& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\

& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\

\end{align}$

Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.

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