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# A fair die is tossed repeatedly. Mr. A wins if it is 1 or 2 and Mr. B wins if it is 3,4,5 or 6. The probability that A wins if the die is tossed indefinitely is:

Hint: Probability is a good topic of mathematics dealing with chances for some events to occur. This question has a fair die to be tossed. We have to find the probability of someone, based on repeated toss. A and B both will have alternate chances to toss till someone wins.

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 $\dfrac{5}{{21}}$.

Note: Here, we have used additive and multiplicative both principles. Additive for happening at one of the many events. And multiplicative for happenings of all events together. Also the formula for the sum of infinite terms of a geometric progression means GP is used here.
Probability is the possible outcomes divided by the total number of outcomes.