Question

A cash register contains 5 $10 bills and 3 $5 bills. you randomly pick 4 bills.

Answer 1

a. The probability of picking exactly two $10 bills is $\frac{3}{7}$

b. The probability of picking at most one $5 bill is $\frac{1}{2}$

Explanation

Number of $10 bills $= 5$

and number of $5 bills $= 3$

Total number of bills= 5+3 = 8. You need to randomly pick 4 bills, so the total possible outcome $= ^8C^4$

a.

For getting exactly two $10 bills, you need to pick two $10 bills and two $5 bills. That means we will pick two $10 bill from 5 bills and two $5 bills from 3 bills.

So, the probability $=\frac{(^5C^2)*(^3C^2)}{^8C^4} =\frac{10*3}{70}=\frac{30}{70}=\frac{3}{7}$

b.

For getting at most one $5 bill, you need to pick either zero or one $5 bill.

If you pick zero  $5 bill, that means there are  four  $10 bills

and if you pick one  $5 bill, that means there are three  $10 bills.

So, the probability $=\frac{(^3C^0*^5C^4)+(^3C^1*^5C^3)}{^8C^4}=\frac{(1*5)+(3*10)}{70}=\frac{5+30}{70}=\frac{35}{70}=\frac{1}{2}$