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There is a 4 digit number to be formed. There are numbers from 0 to 9 in one bag. Digits are pulled out from the bag one after the other to form the number. What is the probability that the 4 digit number is divisible by 9?
  • a)
    13/28
  • b)
    117/252
  • c)
    13/252
  • d)
    117/28
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
There is a 4 digit number to be formed. There are numbers from 0 to 9...
For a number to be divisible by 9, the sum of its digits has to be 9. Let the number be abcd. Then a + b + c + d = 9K
In the given scenario, a + b + c + d can be 9, 18 or 27 only
Total Numbers that can be formed = 9 × 9 × 7 × 8 = 4536 ( 0 cannot be first digit)
WE solve the question by looking at possible cases
Case 1 : Sum of digits is 9
Possible combinations are = {(4,3,2,0)}
4 digits numbers that can be formed = 18
Case 2 : Sum of digits is 18
Possible combinations are =
{(0,9,8,1)(0,9,7,2)(0,9,6,3)(0,9,5,4)(1,9,7,2)(1,9,5,3),(2,9,4,3)}
4 digits numbers that can be formed = 18×4 + 24×3 = 144
Case 3 : Sum of digits is 27
Possible combinations are = {(9,8,7,3)(9,8,6,4)(9,7,6,5))}
4 digits numbers that can be formed = 24×3 = 72
Total Possible numbers = 234
Probability = 234/4536 = 13/252
Hence, the correct option is (c).
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There is a 4 digit number to be formed. There are numbers from 0 to 9 in one bag. Digits are pulled out from the bag one after the other to form the number. What is the probability that the 4 digit number is divisible by 9?a)13/28b)117/252c)13/252d)117/28Correct answer is option 'C'. Can you explain this answer?
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