What is the probability that a randomly selected integer from 100,000 to 999,999 contains exactly five occurrences of the digit 3?

To determine the probability that a randomly selected integer from 100,000 to 999,999 contains exactly five occurrences of the digit 3, we first need to calculate the total number of integers in that range and then find how many of those integers fit the criteria of having exactly five threes.

The range from 100,000 to 999,999 includes all six-digit integers. There are a total of 999,999 - 100,000 + 1 = 900,000 integers in this range.

Next, we focus on finding how many six-digit integers contain exactly five occurrences of the digit 3. If five of the six digits must be 3, there is one remaining digit that must not be 3.

1. Choosing Positions for 3s: There are 6 positions in a six-digit number, and we need to select 5 of these positions to place the digit 3. The number of ways to choose 5 out of 6 is given by the combination formula C(6, 5), which is equal to 6.

2. Choosing the Last Digit: The last remaining position can contain any digit from 0 to 9, except for 3