Answer :

Given that A,B and C throws a die.

The first who throw 6 wins the game.

⇒ P(S_{6}) = P(getting 6)

⇒

⇒ P(S_{N}) = P(not getting 6)

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⇒

Let us assume A starts the game, A wins the game only when he gets 6 while throwing dice in 1^{st},4^{th},7^{th},…… times

Here the probability of getting sum 6 on throwing a dice is same for the players A, B and C

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(A_{wins}) = P(S_{9}) + P(S_{N})P(S_{N})P(S_{N})P(S_{9}) + P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{9}) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

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B wins the game only when he gets 6 while throwing dice in 2^{nd},5^{th},8^{th},…… times and others doesn’t get 6.

Since throwing a dice is an independent event, their probabilities multiply each other

⇒ P(B_{wins}) = (P(S_{N})P(S_{9})) + (P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{9})) + (P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{N})P(S_{9})) + ……………

⇒

⇒

The series in the brackets resembles the Infinite geometric series.

We know that sum of a infinite geometric series with first term ‘a’ and common ratio ‘o’ is .

⇒

⇒

⇒

⇒

⇒ P(C_{wins}) = 1-P(A_{wins})-P(B_{wins})

⇒

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∴ The probabilities of winning of A, B and C is .

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