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Getting 1 or 2 by A will have probability P[A]= $\dfrac{2}{6} = \dfrac{1}{3}$

Getting 3,4,5 or 6 by B will have probability P[B]=$\dfrac{4}{6} = \dfrac{2}{3}$

Thus probability of group AB will be P[AB] = $\dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{2}{9}$

Let us assume that A win the first round, then for A to win, the possible combinations will be:

AA, ABAA, ABABAA, and so on.

Now, the Probability of win of A will be:

\[ = {\text{ }}P\left[ {AA} \right] + P\left[ {ABAA} \right] + P\left[ {ABABAA} \right] + \ldots \ldots .\]

Thus getting probability of each term,

\[

P[A] \times P[A] + P[A] \times P[B] \times P[A] \times P[A] + P[A] \times P[B] \times P[A] \times P[B] \times P[A] \times P[A] + .... \\

\Rightarrow \dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} + .... \\

\Rightarrow \dfrac{1}{9} + \dfrac{2}{{81}} + \dfrac{4}{{729}} + .... \\

\]

It is a GP series with infinite terms.

Formula for its sum is,

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

Here first term , a = $\dfrac{1}{9}$

Common ration, r=$\dfrac{2}{9}$

Therefore, sum will be

$

\dfrac{{\dfrac{1}{9}}}{{1 - \dfrac{2}{9}}} \\

= \dfrac{1}{7} \\

$

Similarly, let us assume that B win the first round, then for A to win, the possible combinations will be:

BAA, BABAA, BABABAA and so on.

Now, the Probability of win of A will be the total of probability of each term.

Thus,

\[

= P\left[ {BAA} \right] + P\left[ {BABAA} \right] + P\left[ {BABABAA} \right] + \ldots \ldots . \\

\Rightarrow P[B] \times P[A] \times P[A] + P[B] \times P[A] \times P[B] \times P[A] \times P[A] + P[B] \times P[A] \times P[B] \times P[A] \times P[B] \times P[A] \times P[A] + .... \\

\Rightarrow \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} \times \dfrac{1}{3}.... \\

\Rightarrow \dfrac{2}{{27}} + \dfrac{4}{{729}} + \dfrac{8}{{6561}} + .... \\

\] It is a GP series with infinite terms.

Formula for its sum is,

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

Here first term , a = $\dfrac{2}{{27}}$

Common ration, r=$\dfrac{2}{9}$

Therefore, sum will be

= $

\dfrac{{\dfrac{2}{{27}}}}{{1 - \dfrac{2}{9}}} \\

= \dfrac{2}{{21}} \\

$

Here one of the above two cases will happen. So, by the additive principle, total probability for win of A, will be:

$

= \dfrac{1}{7} + \dfrac{2}{{21}} \\

= \dfrac{5}{{21}} \\

$

So, the probability of win of A will be $\dfrac{5}{{21}}$.

Probability is the possible outcomes divided by the total number of outcomes.