8 attempts remaining.
(1 point) Rework problem 31 from section 2.4 of your text involving the selection of football plays from a playbook. Assume that the playbook contains 10 passing plays and 15 running plays. The coach randomly selects 8 plays from the playbook.
What is the probability that the coach selects at least 2 passing plays and at least 3 running plays?
(Note: The textbook's version of this problem was solved on The Finite Show. You can view the streaming video of this solution at TFS solution.

Step 1 ：The problem is asking for the probability of selecting at least 2 passing plays and at least 3 running plays from a total of 8 plays selected from a playbook of 10 passing plays and 15 running plays.

Step 2 ：To solve this, we need to calculate the total number of ways to select 8 plays from the playbook, and the number of ways to select at least 2 passing plays and at least 3 running plays.

Step 3 ：The total number of ways to select 8 plays from the playbook is given by the combination formula \(C(n, r) = \frac{n!}{(n-r)!r!}\), where n is the total number of items (in this case, the total number of plays in the playbook), and r is the number of items to be selected (in this case, 8 plays).

Step 4 ：The number of ways to select at least 2 passing plays and at least 3 running plays can be calculated by summing the number of ways to select 2 passing plays and 6 running plays, 3 passing plays and 5 running plays, 4 passing plays and 4 running plays, 5 passing plays and 3 running plays, 6 passing plays and 2 running plays, 7 passing plays and 1 running play, and 8 passing plays and 0 running plays.

Step 5 ：The probability is then given by the ratio of the number of ways to select at least 2 passing plays and at least 3 running plays to the total number of ways to select 8 plays from the playbook.

Step 6 ：Final Answer: The probability that the coach selects at least 2 passing plays and at least 3 running plays is approximately \(\boxed{0.9346}\).