A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 is :

Question

A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 is : (A) 23/500 (B) 11/200 (C) 7/100 (D) none

Tardigrade Admin · Accepted Answer

n ( S )=1000 ; 000,001, ldots ldots . .999 n ( A )= text coeff of x 9 text in (1+ x + x 2+ ldots .+ x 9)3 = text coeff of x 9 text in (1- x 10)3(1- x )-3 = 11 C 9=55 P ( E )= (55/1000)=(11/200)

Q. A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 is :

23/500

11/200

7/100

none

$n (S) = 1000; {000, 001, \dots\dots ..999}$
$n (A) = coeff of x^{9} in (1 + x + x^{2} + \dots . + x^{9})^{3}$
$= coeff of x^{9} in (1 - x^{10})^{3} (1 - x)^{- 3}$
$=^{11} C_{9} = 55$
$P (E) = \frac{55}{1000} = \frac{11}{200}$

Q. A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 is :

23/500

11/200

7/100

none

n(S)=1000;{000,001,……..999} n(A)= coeff of x9 in (1+x+x2+….+x9)3 = coeff of x9 in (1−x10)3(1−x)−3 =11C9​=55 P(E)=100055​=20011​

$n (S) = 1000; {000, 001, \dots\dots ..999}$
$n (A) = coeff of x^{9} in (1 + x + x^{2} + \dots . + x^{9})^{3}$
$= coeff of x^{9} in (1 - x^{10})^{3} (1 - x)^{- 3}$
$=^{11} C_{9} = 55$
$P (E) = \frac{55}{1000} = \frac{11}{200}$