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**Hint:**Find the probability when each person wins and loses. We consider infinite rounds of rolling the dice because it has not been mentioned in the question. Use the following pattern to obtain the equation.

Consider A as the event A winning(A’ as A not winning) and B as the event of B winning(B’ as B not winning).

After round ‘n’ (if both A and B fails to win in round (n-1)$^{th}$ round),

\[\begin{align}

& P(B\,winning) \\

& =\underbrace{\underbrace{\underbrace{\left( P\left( A' \right)\times P\left( B \right) \right)}_{round\,1}+\left( P\left( A' \right)\times P\left( B' \right)\times P\left( A' \right)\times P(B) \right)}_{round\,2}+\left( P\left( A' \right)\times P\left( B' \right)\times P\left( A' \right)\times P(B')\times P\left( A' \right)\times P\left( B \right) \right)}_{round\,3}+......... \\

\end{align}\]

The sum of an infinite geometric progression having its common ratio (r) less than one and first term ‘a’ is,

${{S}_{\infty }}=\dfrac{a}{1-r}$

**Complete step by step answer:**

We are given that A and B throw a pair of dice alternately and that A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10.

We are asked to find the probability that B wins.

Let A be the event that A wins, that is, A wins (A gets a total of 7). Let B be the event that B wins (B gets a total of 10).

We know that A wins when the sum of the numbers on the two dice equals to 7. Hence, the different possibilities when A will win is if (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3) are the values obtained on rolling the dice.

Mathematically speaking,

A = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

These are the cases in which A will win.

We know that B wins when the sum of the numbers on the two dice equals to 10. Hence, the different possibilities when B will win is if (4, 6) (6, 4) and (5, 5) are the values obtained on rolling the dice.

Mathematically speaking,

B = {(4, 6) (6, 4), (5, 5)}

These are the cases in which B will win.

We know that the total number of outcomes (Also called sample space) of rolling a pair of dice is 36.

Hence,

Probability of event A, \[P(A)=\dfrac{No.of\,outcomes\,in\,A}{Sample\,space}=\dfrac{6}{36}=\dfrac{1}{6}\,.............(1)\]

Probability of event B, \[P(B)=\dfrac{No.of\,outcomes\,in\,B}{Sample\,space}=\dfrac{3}{36}=\dfrac{1}{12}\,................(2)\]

We are given that person A starts the game. Person B rolls the dice only if A loses.

We are asked to find the probability of person B winning. Person B can win if and only if person A fails to obtain 7.

Probability of person A not winning, \[P(A')=1-P(A)=\dfrac{30}{36}=\dfrac{5}{6}\,.............(3)\]

Probability of person B not winning, \[P(B')=1-P(B)=\dfrac{33}{36}=\dfrac{11}{12}\,.............(4)\]

If both A and B fail to win on their first throws, they continue rolling the dice until one of them wins. But we only need to consider the cases in which B wins.

Hence, \[P(B)=P(B\ wins\ on\ first\ throw)+P(B\ wins\ on\ second\ throw)+P(B\ wins\ on\ third\ throw)....\] \[=P({A}'B)+P({A}'{B}'{A}'B)+P({A}'{B}'{A}'{B}'{A}'B)....\]

\[=\dfrac{5}{6}\times \dfrac{1}{12}+\dfrac{5}{6}\times \dfrac{11}{12}\times \dfrac{5}{6}\times \dfrac{1}{12}+\dfrac{5}{6}\times \dfrac{11}{12}\times \dfrac{5}{6}\times \dfrac{11}{12}\times \dfrac{5}{6}\times \dfrac{1}{12}....\]

\[=\dfrac{5}{72}+\dfrac{5}{72}\times \dfrac{55}{72}+\dfrac{5}{72}\times \dfrac{55}{72}\times \dfrac{55}{72}....\]

\[=\dfrac{5}{72}\left[ 1+\dfrac{55}{72}+{{\left( \dfrac{55}{72} \right)}^{2}}+.... \right]\]

We can see that the RHS is in the form of the sum of an infinite geometric progression.

We know that, the sum of an infinite geometric progression having its common ratio (r) less than one and first term ‘a’ is,

${{S}_{\infty }}=\dfrac{a}{1-r}$

In our case, $a=1$ and $r=\dfrac{55}{72}$

\[\therefore P(B)=\dfrac{5}{72}\left( \dfrac{1}{1-\dfrac{55}{72}} \right)=\dfrac{5}{17}\]

**Hence, the probability of B winning is \[\dfrac{5}{17}\] .**

**Note:**This question could have been asked for a specific number of rounds (Rather than infinite). Say it was asked for 10 rounds.

Then we have to use the following equation for sum of geometric progression,

${{S}_{10}}=\dfrac{a(1-{{r}^{10}})}{1-r}$

The number of events can also be different. They could have introduced a person C, which would give a total of 6 events (Person A,B,C each winning or losing).

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