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 = 1135/1092
= 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.
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