Twelve players S1, S2,...,S12 play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that at least one of S1 and S2 is among the six winners is:

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Question

Twelve players S₁, S₂,...,S₁₂ play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that at least one of S₁ and S₂ is among the six winners is:

$\frac{32}{33}$

$\frac{10}{11}$

$\frac{31}{33}$

None of these

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given: S₁, S₂,…S₁₂ are 12 players play in chess tournament.
Let P(exactly one of S₁ and S₂ is among the six winners)

= P₁(atleast one of S₁ and S₂ is among the six winner)

– P₂ (Both S₁ and S₂ are among the six winners)

⇒ P₁(atleast one of S₁ and S₂ is among the six winner)
= P(exactly one of S₁ and S₂ is among the six winners)
+ P₂(Both S₁ and S₂ are among the six winners)

$= \frac{14}{33} + \frac{6}{11}$

⇒ $p_{1} = \frac{32}{33}$

P₁ (Alteast one S₁ and S₂is among the six winners) $= \frac{32}{33}$