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Grade 11Algebra

Sixteen players S1, S₂ . . . . . . . . . . . . . . . . . . . S₁₆ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength
(a) Find the probability that the player S₁ is among the eight winners.
(b) Find the probability that exactly one of the two players S₁ and S₂ is among the eight winners.

Simran Bhatia

12 Years agoGrade 11

2 Answers

Aditi Chauhan

12 Years ago

Hello Student,

Please find the answer to your question

(a) Prob. of S₁ to among the eight winners = (prob. of S₁ being in a pair) x (prob. of S₁ winning in the group).

= 1 x 1/2 [∵ S₁ is definitely in a group i.e. certain event] = 1/2

(b) If S₁ and S₂ are in the same pair then exactly on wins. If S₁ and S₂ are in two pairs separately then exactly one of S₁ and S₂ will be among the eight winners if S₁ win and S₂ loses or S₁loses and S₂ wins.

Now the prob. of S₁, S₂ being in the same pair and one wins.

= (prob. of S₁, S₂ being in the same pair) x (prob. of any one winning in the pair)

And the prob. of S₁, S₂ being in the same pair

= n (E)/n (S), where n (S) = the no. of ways in which 16 person can be divided in 8 pairs; n (E) = the no. of ways in which S₁, S₂ are in same pair or 14 persons can be divided into 7 pairs.

∴ n (E) = 14!/(2!)⁷ .7! and n (S) = 16!/(2!)⁸ .8!

∴ Prob. of S₁ and S₂ being in the same pair

= $\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}$ = 21.8/16.15 = 1/15

The prob. of any one winning in the pair of S₁, S₂ = P (certain event) = 1

∴ The pair of S₁, S₂ being in two pairs separately and any one of S₁, S₂ wins.

= the prob. of S₁, S₂ being in two pairs separately and S₁ wins, S₂ loses + the prob. of S₁, S₂ being in two pairs separately and S₁ loses, S₂ wins.

= $\left [ 1-\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}\right]$ x 1/2 x 1/2 + $\left [ 1-\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}\right]$ x 1/2 x 1/2

=

x 1/4 = 1/2 x 14 x 14!/15 x 14! = 7/15

∴ Required prob. = 1/15 + 7/15 = 8/15

Thanks
Aditi Chauhan
askIITians Faculty

Sher Mohammad

12 Years ago

Since the probablity of winning each is is ½, so the probablity of s1 win is ½.

the second probablity is ( 1/15 ) + ( 14/15 ) ( ½ ) = 8/15

sher mohammad

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