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 2nd throw or 4th or 6th 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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