Question

Assume that there are 16 board members: 10 females, and 6 males including Larry. There are 4 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment.

(1) Find the probability that both males and females are given a task.

(2) Find the probability that Larry and at least one female are given tasks.

Answer 1

(1) That would be 1-(prob. of no males)- (prob. of no females).
1 - (10*9*8*7)/(16*15*14*13)
- (6*5*4*3)/(16*15*14*13)
= 1 - (5040 + 360)/43680
= 1 - 5400/43680 = 1-135/1092
= 957/1092 = 319/364

2) Larry will be picked in
1 - (15/16)*(14/15)*(13*14)*(12/13)
= 1 - 32760/43680 = 1- 819/1092
= 273/1092 of the situations.
Next, calculate the fraction of those situations for which the three other tasks are assigned to one or more women. Multiply 273/1092 by that factor.