BITSAT Mathematics Question (2024) — Solution

Step-by-step solution

n(S)= Total number of ways =1000 The favourable cases that the sum of the digits of the marked number on the page is equal to 9 are one digit number or two digits numbers or three digits numbers, if three digit number is a b c. Then, a+b+c=9 ; 0 a, b, c 9. n(E)= Number of favourable ways = Number of solutions of the equation = ^ 9+3-1 C_ 3-1 = ^ 11 C_2=55 Required probability = n(E) n(S) = 55 1000 = 11 200 .