wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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. The probability that "exactly one of the two players s1 & s2 is among the eight winners" is

A
415
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
715
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
815
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
915
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 815
Total number of ways of selecting winners =16C8
favourable ways =2C1.14C7 (out of s1 & s2 & rest)
so probability =2C114C716C8=815

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Binomial Experiment
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon