What is the probability that the coach selects at least 2 passing plays and at least 3 running plays?

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?|||answer = probability that he selects exactly 2, 3, 4 or 5 passing plays





prob that he selects exactly two passing plays =


(8C2)*(11*10)*


(15*14*13*12*11*10)


/(26*25*...19)





where:


8C2 is the number of ways of choosing two slots out of 8


and prob that he selects first passing play = 11/26


prob that he selects second passing play = 10/25 etc





You can calculate the other three cases similarly and sum to obtain the answer.|||low|||Total number of selections is 26 C 8 = 1562275





Then, to meet the criteria, he can choose


2 passing, 6 running


3 passing, 5 running


4 passing, 4 running


5 passing, 3 running





The total of all those is:


11C2 * 15C6 + 11C3 * 15C5 + 11C4 * 15C4 + 11C5 * 15C3 which is


1431430





Making the probability: 1431430 / 1562275 which = 91.62 %





It's not much work to check the complement of that in the same way:





11C0 * 15C8 + 11C1 * 15C7 + 11C6 * 15C2 + 11C7 * 15C1 + 11C8 * 15C0


= 130845


130845 / 1562275 = 8.38 % of selecting too few or too many running plays.





130845 + 1431430 = 1562275 which is what we want