Question 1122537: A bag contains two 1-dollar bills, three 5-dollar bills, and two 10-dollar bills. An experiment consists of drawing three random bills from the bag simultaneously. Considering an outcome to be the three selected bills, the experiment has C(7,3)=35 equally likely outcomes in its sample space. Let E be the event that at most one 5-dollar bill was drawn. What is n(E)?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
n(E) is the number of ways of choosing either none of the three $5 bills AND 3 of the four others, OR one of the three $5 bills AND two of the four others:
C(3,0)*C(4,3)+C(3,1)*C(4,2) = 1*4+3*6 = 4+18 = 22.
If you are just learning how to solve problems like this, notice that the ANDs indicate multiplications and the ORs indicate additions.
When learning to solve problems like this, it is good practice to verify that those possibilities plus the others give the correct total of 35 total ways to choose 3 of the 7 bills.
two $5 bills and one of the others: C(3,2)*C(4,1) = 3*4 = 12
three $5 bills and none of the others: C(3,3)*C(4,0) = 1*1 = 1
22+12+1 = 35 CHECK!
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