One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM

DETAILS

MRP

DISCOUNT

FINAL PRICE

Total Price: Rs.

There are no items in this cart.

Continue Shopping

Continue Shopping

Sixteen players S1, S_{2} . . . . . . . . . . . . . . . . . . . S_{16} 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 S_{1} is among the eight winners.

(b) Find the probability that exactly one of the two players S_{1} and S_{2} is among the eight winners.

6 years ago

Hello Student,

Please find the answer to your question

(a) Prob. of S_{1} to among the eight winners = (prob. of S_{1} being in a pair) x (prob. of S_{1} winning in the group).

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

(b) If S_{1} and S_{2} are in the same pair then exactly on wins. If S_{1} and S_{2} are in two pairs separately then exactly one of S_{1} and S_{2} will be among the eight winners if S_{1} win and S_{2} loses or S_{1 }loses and S_{2} wins.

Now the prob. of S_{1}, S_{2} being in the same pair and one wins.

= (prob. of S_{1}, S_{2} being in the same pair) x (prob. of any one winning in the pair)

And the prob. of S_{1}, S_{2} 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 S_{1}, S_{2} 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 S_{1} and S_{2} being in the same pair

= = 21.8/16.15 = 1/15

The prob. of any one winning in the pair of S_{1}, S_{2} = P (certain event) = 1

∴ The pair of S_{1}, S_{2} being in two pairs separately and any one of S_{1}, S_{2} wins.

= the prob. of S_{1}, S_{2} being in two pairs separately and S_{1} wins, S_{2} loses + the prob. of S_{1}, S_{2} being in two pairs separately and S_{1} loses, S_{2} wins.

= x 1/2 x 1/2 + 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

6 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

faculty askiitians

b.tech, iit delhi

copyright © 2006-2021 askIITians.com

info@askiitians.com