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 that the coach selects at least 3 passing plays and at least 2 running plays, we need to calculate the probability of two separate events: selecting at least 3 passing plays and selecting at least 2 running plays.
Let's begin by calculating the probability of selecting at least 3 passing plays.
The total number of ways to choose 8 plays from a playbook of 10 passing plays is given by the combination formula C(10, 8). This can be calculated as follows:
C(10, 8) = 10! / (8!(10-8)!) = (10 * 9) / (2 * 1) = 45
Next, we need to consider the number of ways to choose fewer than 3 passing plays.
The number of ways to choose 0 passing plays is C(10, 0) = 1.
The number of ways to choose 1 passing play is C(10, 1) = 10.
The number of ways to choose 2 passing plays is C(10, 2) = 45.
Therefore, the probability of selecting fewer than 3 passing plays is (1 + 10 + 45) / 45 = 56 / 45.
Now let's calculate the probability of selecting at least 2 running plays.
The total number of ways to choose 8 plays from a playbook of 12 running plays is given by the combination formula C(12, 8). This can be calculated as follows:
C(12, 8) = 12! / (8!(12-8)!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495.
Next, we need to consider the number of ways to choose fewer than 2 running plays.
The number of ways to choose 0 running plays is C(12, 0) = 1.
The number of ways to choose 1 running play is C(12, 1) = 12.
Therefore, the probability of selecting fewer than 2 running plays is (1 + 12) / 495 = 13 / 495.
To find the probability of selecting at least 3 passing plays and at least 2 running plays, we can use the principle of inclusion-exclusion. We can subtract the probability of selecting fewer than 3 passing plays or fewer than 2 running plays.
P(at least 3 passing plays and at least 2 running plays) = 1 - [P(fewer than 3 passing plays) + P(fewer than 2 running plays) - P(fewer than 3 passing plays and fewer than 2 running plays)]
P(at least 3 passing plays and at least 2 running plays) = 1 - [56/45 + 13/495 - P(fewer than 3 passing plays and fewer than 2 running plays)]
To calculate P(fewer than 3 passing plays and fewer than 2 running plays), we need to determine the number of ways to choose fewer than 3 passing plays and fewer than 2 running plays simultaneously.
The number of ways to choose 0 passing plays and 0 running plays is C(10, 0) * C(12, 0) = 1 * 1 = 1.
The number of ways to choose 0 passing plays and 1 running play is C(10, 0) * C(12, 1) = 1 * 12 = 12.
The number of ways to choose 1 passing play and 0 running plays is C(10, 1) * C(12, 0) = 10 * 1 = 10.
The number of ways to choose 1 passing play and 1 running play is C(10, 1) * C(12, 1) = 10 * 12 = 120.
The number of ways to choose 2 passing plays and 0 running plays is C(10, 2) * C(12, 0) = 45 * 1 = 45.
The number of ways to choose 0 passing plays and 2 running plays is C(10, 0) * C(12, 2) = 1 * 66 = 66.
Therefore, P(fewer than 3 passing plays and fewer than 2 running plays) = (1 + 12 + 10 + 120 + 45 + 66) / 45 / 495 = 254 / 495.
Substituting this value back into the calculation:
P(at least 3 passing plays and at least 2 running plays) = 1 - (56/45 + 13/495 - 254/495) = 453 / 495.
So, the probability that the coach selects at least 3 passing plays and at least 2 running plays is 453/495.