3. (7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket where the digits are in ascending (descending) order?
Solution: There are a total of $10^{6}$ different tickets, of which $C_{10}^{6}$ have digits in ascending (or descending) order (any 6 digits out of 10 form exactly one number we need). Then the probability is $\frac{C_{10}^{6}}{10^{6}}=0.00021$.

(7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket with all different digits?

Solution: There are a total of $10^{6}$ different tickets, of which $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5$ tickets have all different digits (the first digit can be chosen in 10 ways, the second in 9 ways, and so on). Then the probability is $\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{10^{6}}=0.1512$.

(7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket where any a) two; b) three; c) four consecutive digits are different?

Solution: There are a total of $10^{6}$ different tickets, of which a) $10 \cdot 9^{5}$; b) $10 \cdot 9 \cdot 8^{4}$; c) $10 \cdot 9 \cdot 8 \cdot 7^{3}$ have any a) two; b) three; c) four consecutive digits different. Then the probability is a) $\frac{10 \cdot 9^{5}}{10^{6}}=0.59049$; b) $\frac{10 \cdot 9 \cdot 8^{4}}{10^{6}}=0.36864$; c) $\frac{10 \cdot 9 \cdot 8 \cdot 7^{3}}{10^{6}}=0.24696$.

(7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket with at least two identical digits?

Solution: The number of tickets with at least two identical digits is $10^{6}-10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5$ (from all tickets, subtract those with all different digits). Then the probability is $\frac{10^{6}-10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{10^{6}}=0.8488$.

(7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket with at least one digit a) greater than 5; b) less than 3?

Solution: The number of tickets with at least one digit a) greater than 5; b) less than 3, is a) $10^{6}-6^{6}$; b) $10^{6}-7^{6}$ (from all tickets, subtract those where all digits a) do not exceed 5; b) are not less than 3). Then the probability is a) $\frac{10^{6}-6^{6}}{10^{6}}=0.953344$; b) $\frac{10^{6}-7^{6}}{10^{6}}=0.882351$.

(7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket with exactly one digit less than 3?

Solution: The number of tickets with exactly one digit less than 3 is $6 \cdot 3 \cdot 7^{5}$ (6 ways to choose the position where the digit less than 3 will stand, and on the remaining 5 positions, the digits are not less than 3). Then the probability is $\frac{6 \cdot 3 \cdot 7^{5}}{10^{6}}=0.302526$.

(7 points) Lottery tickets are numbered with five digits: from 00000 to 99999. You buy one ticket. What is the probability that you will get a ticket with at least one even digit?

Solution: The number of tickets with at least one even digit is $10^{5}-5^{5}$ (from all tickets, subtract those with all digits odd). Then the probability is $\frac{10^{5}-5^{5}}{10^{5}}=0.96875$.