number of task assignments without restrictions = C(17,4) = 2380
Carl ---- C
Female --F
Male ----M
cases:
CFMM --- 1*C(9,1)*C(7,2) = 189
CFFM ----1*C(9,2)*C(7,1) = 252
CFFF -----1*C(9,3)*C(7,0) = 84
total cases with Carl and at least one female = 525
prob of the event as stated = 525/2380 = 15/68
Assume that there are 17 board
members: 9 females, and 8 males including Carl. There are 4 tasks to be assigned.
What is the probability that at least one female and Carl are assigned tasks?
1 answer