CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Three natural numbers are taken at random from the set A={x|1x100,xϵN}. The probability that the AM of the numbers is 25, is

A
77C2100C3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
25C2100C3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
74C72100C97
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 74C72100C97
n(S)=100C3
Since the A.M. of three numbers is 25.
Therefore their sum = 75
n(E)= the number of integral solution of x1+x2+x3=75
Where x11,x21,x31
= coeff. of x75 in (x+x2+x3+...)3
=coeff. of x72 in (1+x+x2+...)3
=coeff. of x72 in (11x)3
=coeff. of x72 in (1x)3
=74C72=74C2
P(E)=74C2100C3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Addition rule
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon