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A and B alternately throw a pair of die until one of them wins. A is considered winner if he gets a total 7 and B is considered winner, if he gets a total of 5. If A starts the game. then what is the probability that one of the die shows face 3 on the last throw.

I am not getting any start . Can anybody help me in this .

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    First you need to find the probability of A (B) wins in a single throw. Then you can calculate the probability to win in the game by summing probabilities. After that use total probability law with two hypotheses: {A wins} or {B wins} to find the probability that face 3 appears. The answer 9/28 is not true. – 2017-02-19

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First off: The chance that A wins in any single attempt is 6/36, i.e. 1/6 - if you write out all possible results for a 2-dice-throw, this gets obvious:
(1,6), (6,1), (2,5), (5,2), (3,4), (4,3) are possible wins for A.

Chance to win for B is 4/36, i.e. 1/9:
(1,4), (4,1), (2,3), (3,2)

The chance that there is a three in the result is 1/3 for A and 1/2 for B.

With this knowledge you can now build an infinite sum:
Possibility 1: A wins and throws 3: $\frac16*\frac13$
Possibility 2: A loses, B wins and throws 3: $\frac56*\frac19*\frac12$
Possibility 3: A loses, B loses, A wins and throws 3: $\frac 56*\frac89*\frac16*\frac13$
Possibility 4: ... : $\frac56*\frac89*\frac56*\frac19*\frac12$
...

So the total sum is: $$\frac1{18}+\frac5{108}+\sum_{i=1}^\infty\left(\frac56*\frac89\right)^i*\frac 1{18}+\sum_{j=1}^\infty\left(\frac56*\frac89\right)^j*\frac56*\frac1{18}=$$ $$\frac1{18}+\frac5{108}+\frac{20}7*\frac1{18}+\frac{20}7*\frac56*\frac1{18}=\frac{11}{28}$$

Are you sure that the result is 9/28?

EDIT: (I can't answer your comment directly, as I'm a new member). The sums should be correct, write the first few summands out, if you have doubts.

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    http://dc.allenbpms.in/testpaper/solution/d8947-79-636229541612684899-079s.gif – 2017-02-19
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    Both values of i and j contains only odd numbers – 2017-02-19