13. A box contains tickets numbered 1 through 10. A person draws two tickets from the box without replacement. What is the probability of: a. Drawing a ticket with a number less than 5 and a ticket with a number greater than 8 ? b. Drawing two tickets with even numbers? c. Drawing a ticket with an odd number followed by a ticket with an even number?

To find the probabilities for each case, let's break them down one by one. **a. Probability of drawing a ticket with a number less than 5 and a ticket with a number greater than 8:** First, we identify the tickets less than 5, which are: 1, 2, 3, and 4 (4 tickets total). The tickets greater than 8 are: 9 and 10 (2 tickets total). When drawing two tickets without replacement, we can have two scenarios: 1. Draw one ticket less than 5 first and then one ticket greater than 8. 2. Draw one ticket greater than 8 first and then one ticket less than 5. Calculating for the first scenario: - Probability of the first ticket being less than 5: \( \frac{4}{10} \) - After drawing that ticket (let's say one of the numbers less than 5), we have 9 tickets left, one of which is greater than 8. The probability then becomes \( \frac{2}{9} \). So for this scenario, the combined probability is: \[ P(\text{less than 5 first}) = \frac{4}{10} \times \frac{2}{9} = \frac{8}{90} = \frac{4}{45} \] Now for the second scenario: - Probability of the first ticket being greater than 8: \( \frac{2}{10} \) - Now we have 9 tickets left, and the probability of picking a ticket less than 5 is \( \frac{4}{9} \). The combined probability for this scenario is: \[ P(\text{greater than 8 first}) = \frac{2}{10} \times \frac{4}{9} = \frac{8}{90} = \frac{4}{45} \] Finally, we sum both scenarios: \[ P = \frac{4}{45} + \frac{4}{45} = \frac{8}{45} \] **b. Probability of drawing two tickets with even numbers:** The even numbered tickets in the range are: 2, 4, 6, 8, and 10 (5 tickets total). Now we find the probability of drawing two even tickets: - The probability of the first ticket being even: \( \frac{5}{10} = \frac{1}{2} \) - After drawing one even ticket, we have 4 even tickets left out of a total of 9 remaining tickets, so the probability of the second ticket being even is \( \frac{4}{9} \). Thus, the combined probability of drawing two even tickets is: \[ P = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9} \] **c. Probability of drawing a ticket with an odd number followed by a ticket with an even number:** The odd numbered tickets in the range are: 1, 3, 5, 7, and 9 (5 tickets total). The even numbered tickets are still 2, 4, 6, 8, and 10 (5 tickets total). Now we find the probability for this sequence: - The probability of the first ticket being odd: \( \frac{5}{10} = \frac{1}{2} \) - After drawing one odd ticket, we have 5 even tickets still available out of 9 remaining tickets, so the probability of the second ticket being even is \( \frac{5}{9} \). Thus, the combined probability of drawing one odd ticket followed by one even ticket is: \[ P = \frac{1}{2} \times \frac{5}{9} = \frac{5}{18} \] In summary: a. Probability of drawing one ticket with a number less than 5 and one greater than 8 is \( \frac{8}{45} \). b. The probability of drawing two tickets with even numbers is \( \frac{2}{9} \). c. The probability of drawing one odd ticket followed by one even ticket is \( \frac{5}{18} \).