Given that A and B throw a pair of die alternatively.

It is told that A wins if he gets sum of 7 and B wins if he gets sum of 10.

Let us find the probability of getting sum 7 and 10.

The possibilities of getting 7 on throwing a pair of die:

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

The possibilities of getting 10 on throwing a pair of die:

{(4,6),(5,5),(6,4)} = 3 cases

⇒ P(S₇) = P(getting sum 7)

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⇒ P(S_N7) = P(not getting sum 7)

⇒ P(S_N7) = 1-P(S₇)

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⇒ P(S₁₀) = P(getting sum 10)

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⇒ P(S_N10) = P(not getting sum 10)

⇒ P(S_N10) = 1-P(S₁₀)

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It is told that A starts the game.

We need to find the probability that B wins the games.

B wins the game only when A losses in1^st,3^rd,5^th,…… throws.

This can be shown as follows:

⇒ P(B_WINS) = P(B wins the game)

⇒ P(B_Wins) = P(S_N7S₁₀) + P(S_N7S_N10S_N7S₁₀) + P(S_N7S_N10S_N7S_N10S_N7S₁₀) + …………………

Since throwing a pair of die by each person is an independent event, the probabilities multiply each other.

⇒ P(B_wins) = (P(S_N7)P(S₁₀)) + (P(S_N7)P(S_N10)P(S_N7)P(S₁₀)) + ………………

⇒

⇒

The terms in the bracket resembles the infinite geometric series sequence:

We know that the sum of a Infinite geometric series with first term ‘a’ and common ratio ‘r’ is

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∴ The required probability is .