NoviceBoy wrote:
Let C be the collection of all integers between 100,000 and 999,999 inclusive. What is the probability that the number randomly selected from C has exactly 5 occurrences of the digit 3 in the decimal representation?
A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000
Total number of digits between \(999,999\) and \(100,000\), inclusive = \((999,999 - 100,000) + 1 \)
We can infer that each of the numbers between the range is a six-digit number. In that, we want to identify numbers in which the digit \(3\) occurs 5 times. Hence, the sixth digit can be any digit between \(0\) and \(9\), inclusive, except for \(3\). Two cases are possible
1) The number does not start with 3
2) The number starts with 3
Case 1: The number does not start with 3 i.e. the format of the number is \(X33333\)
- If the number doesn't start with three, the first place, \(X\), can be filled by 1, 2, 4, 5, 6, 7, 8, and 9 ⇒ So \(8\) ways
- The next five digits of the number will be filled with 3s
Total = \(8\) ways
Case 2: The number starts with 3
- If the number starts with three, four places of the remaining five places will contain a 3.
- The fifth place can be placed in 9 ways ( 0, 1, 2, 4, 5, 6, 7, 8, and 9)
- We can multiple arrangements of the five digits. This can be done in \(\frac{5!}{4!}\) ways (all possible arrangments of \(3333X\), where \(X\) is a value between 0 and 9, except 3)
Total number of ways = \(9 * \frac{5!}{4!} = 45\)
Favourable cases = \(45 + 8 = 53\)
Total cases = \(900000\)
Required Probability = \(\frac{53}{900,000}\)
What am I missing here?