1. **Problem statement:** We want to find: a. The number of integers from 1 to 100,000 that contain the digit 6 exactly once. b. The number of integers from 1 to 100,000 that contain the digit 6 at least once. c. The probability that a randomly chosen integer from 1 to 100,000 contains two or more occurrences of the digit 6. 2. **Key idea:** We consider all 5-digit numbers from 00001 to 100000 (treating numbers with leading zeros for uniformity). The total count is 100,000. 3. **Counting numbers with exactly one 6:** - Total digits: 5 - Exactly one digit is 6, the other 4 digits are from {0,...,9} excluding 6 (so 9 choices each). - Number of ways to choose the position of the single 6: $\binom{5}{1} = 5$ - For the other 4 positions: $9^4 = 6561$ - Total numbers with exactly one 6: $5 \times 6561 = 32805$ - But we must exclude numbers with leading zeros that are not valid (e.g., 00000 is not in 1 to 100000, but 00006 is 6, which is valid). Since 00000 is not counted, and leading zeros represent smaller numbers, all these counts are valid for numbers 1 to 99999. - For 100000, digit 6 does not appear, so no effect. 4. **Counting numbers with at least one 6:** - Total numbers: 100,000 - Numbers with no 6 at all: each digit from {0,...,9} except 6, so 9 choices per digit - Number of 5-digit strings with no 6: $9^5 = 59049$ - Numbers from 1 to 100000 with no 6: 59049 (including leading zeros representing numbers less than 100000) - So numbers with at least one 6: $100000 - 59049 = 40951$ 5. **Counting numbers with two or more 6's:** - Numbers with exactly one 6: 32805 (from step 3) - Numbers with at least one 6: 40951 (from step 4) - So numbers with two or more 6's: $40951 - 32805 = 9146$ 6. **Probability of two or more 6's:** - Total numbers: 100000 - Probability = $\frac{9146}{100000} = 0.09146$ **Final answers:** a. 32805 b. 40951 c. 0.09146