Posted by **Nala** on Monday, March 2, 2009 at 10:38pm.

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.

- college math -
**drwls**, Tuesday, March 3, 2009 at 12:12am
(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.

- college math -
**Anonymous**, Wednesday, September 14, 2016 at 6:13pm
123

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