Answer :

As captain of team A has started to throw the die, he may win in the first throw or third or fifth and so on.

Team B can win in 2^{nd} throw or 4^{th} or 6^{th} and so on

As probability of getting 6 on a die = 1/6 = P(6)

P(6’) represents event of not getting 6 = 5/6

Let P(A) represents the probability that team A wins the game and P(B) represents the probability that team B wins the game.

∴ P(A) = P(6) + P(6’)P(6’)P(6) + P(6’)P(6’)P(6’)P(6’)P(6)+…to ∞

⇒ P(A) =

⇒ P(A) =

Clearly it forms an infinite GP and sum of infinite GP is given by where a is the first term and r is common ratio.

Here a = 1 and r = (5/6)^{2}

∴ P(A) =

Similarly,

P(B) = P(6’)P(6) + P(6’)P(6’)P(6’)P(6) + … to ∞

⇒ P(B) =

⇒ P(B) =

Clearly it forms an infinite GP and sum of infinite GP is given by where a is the first term and r is common ratio.

Here a = 1 and r = (5/6)^{2}

∴ P(B) =

Clearly, P(A) > P(B)

Hence, there is more chance of winning of team A.

∴ Umpire’s decision is unfair. He has favoured team A.

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