Add up these probabilities:
2 running (5 passing)
1 passing (7 running)
1 running (7 passing)
0 running (8 passing)
0 passing (8 running)
and subtract the total from 1.
The probability of 0 passing plays is (15/26)^8 = 0.01227
You do the others
Assume that the playbook contains 11 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?
1 answer