Assume that the playbook contains 10 passing plays and 12 running plays. The coach randomly selects 8 plays from the playbook.
What is the probability that the coach selects at least 3 passing plays and at least 2 running plays?
To find the probability of the coach selecting at least 3 passing plays and at least 2 running plays, we need to calculate the probability of two scenarios:
1. The coach selects exactly 3 passing plays and 2 running plays, and
2. The coach selects exactly 4 passing plays and 2 running plays, or exactly 5 passing plays and 2 running plays, and so on, up to all 8 plays being passing plays.
Let's calculate the probability for each scenario:
1. The coach selects exactly 3 passing plays and 2 running plays:
The number of ways to choose 3 passing plays from 10 passing plays is given by the combination formula: C(10, 3) = 10! / (3! * (10 - 3)!) = 120.
Similarly, the number of ways to choose 2 running plays from 12 running plays is C(12, 2) = 12! / (2! * (12 - 2)!) = 66.
Therefore, the total number of ways to select exactly 3 passing plays and 2 running plays is 120 * 66 = 7,920.
Now, let's calculate the probability for this scenario:
The total number of ways to choose 8 plays from 22 plays in the playbook is C(22, 8) = 22! / (8! * (22 - 8)!) = 22,957,480.
So, the probability of selecting exactly 3 passing plays and 2 running plays is 7,920 / 22,957,480 ≈ 0.000344.
2. The coach selects exactly 4 passing plays and 2 running plays:
The number of ways to choose 4 passing plays from 10 passing plays is C(10, 4) = 10! / (4! * (10 - 4)!) = 210.
Similarly, the number of ways to choose 2 running plays from 12 running plays is C(12, 2) = 66.
Therefore, the total number of ways to select exactly 4 passing plays and 2 running plays is 210 * 66 = 13,860.
The probability for this scenario is 13,860 / 22,957,480 ≈ 0.000603.
We repeat this calculation for each scenario up to 8 passing plays and 0 running plays, and add up the probabilities for each scenario.
Finally, the probability that the coach selects at least 3 passing plays and at least 2 running plays is the sum of the probabilities for all these scenarios.
To calculate the probability of selecting at least 3 passing plays and at least 2 running plays, we can use the concept of combinations.
First, let's determine the total number of possible combinations the coach can select from the playbook.
The coach has a total of 10 passing plays and 12 running plays in the playbook. Therefore, the total number of combinations that the coach can select is equal to the sum of combinations of selecting 8 plays from passing plays (C(10, 8)) and selecting 8 plays from running plays (C(12, 8)).
To calculate combinations, we use the formula C(n, r) = n! / (r!(n-r)!).
C(10, 8) = 10! / (8!(10-8)!) = (10 * 9) / (2 * 1) = 45
C(12, 8) = 12! / (8!(12-8)!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495
Total number of combinations = C(10, 8) + C(12, 8) = 45 + 495 = 540
Next, let's calculate the number of combinations that satisfy the condition of selecting at least 3 passing plays and at least 2 running plays.
Case 1: Selecting exactly 3 passing plays and exactly 2 running plays.
The coach can choose 3 passing plays in C(10, 3) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 3) * C(12, 2) = (10! / (3!(10-3)!)) * (12! / (2!(12-2)!)) = 120 * 66 = 7920
Case 2: Selecting exactly 4 passing plays and exactly 2 running plays.
The coach can choose 4 passing plays in C(10, 4) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 4) * C(12, 2) = (10! / (4!(10-4)!)) * (12! / (2!(12-2)!)) = 210 * 66 = 13860
Case 3: Selecting exactly 5 passing plays and exactly 2 running plays.
The coach can choose 5 passing plays in C(10, 5) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 5) * C(12, 2) = (10! / (5!(10-5)!)) * (12! / (2!(12-2)!)) = 252 * 66 = 16632
Case 4: Selecting exactly 6 passing plays and exactly 2 running plays.
The coach can choose 6 passing plays in C(10, 6) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 6) * C(12, 2) = (10! / (6!(10-6)!)) * (12! / (2!(12-2)!)) = 210 * 66 = 13860
Case 5: Selecting 7 passing plays and exactly 2 running plays.
The coach can choose 7 passing plays in C(10, 7) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 7) * C(12, 2) = (10! / (7!(10-7)!)) * (12! / (2!(12-2)!)) = 120 * 66 = 7920
Case 6: Selecting all 8 passing plays and exactly 2 running plays.
The coach can choose 8 passing plays in C(10, 8) ways and 2 running plays from C(12, 2) ways.
Number of combinations = C(10, 8) * C(12, 2) = (10! / (8!(10-8)!)) * (12! / (2!(12-2)!)) = 45 * 66 = 2970
Total number of combinations satisfying the condition = Case 1 + Case 2 + Case 3 + Case 4 + Case 5 + Case 6 = 7920 + 13860 + 16632 + 13860 + 7920 + 2970 = 64102
Finally, we can calculate the probability by dividing the number of combinations satisfying the condition by the total number of combinations:
Probability = Number of combinations satisfying the condition / Total number of combinations = 64102 / 540 = 0.1185 (rounded to 4 decimal places)
Therefore, the probability that the coach selects at least 3 passing plays and at least 2 running plays is approximately 0.1185.