8 slips of paper each has a 2 on them; 2 slips of paper each has a 4 of them
The total value of all the slips when added together is (8)(2) + (2)(4) = 16 + 8 = 24
The expected value of one draw is the average value of these 10 slips of paper is 24 / 10 = 2.4
If we add one additional 4, then the total becomes 24 + 4 = 28.
Since there are now 11 slips of paper, the expected value (average) is 28 / 11 = 2.5454...
Adding another 4, the total becomes 28 + 4 = 32.
Since there are now 12 slips of paper, the expected value (average) is 32 / 12 = 2.6666...
How many 4s must be added to get an average of 3?
Starting with the original 10 slips of paper and the original sum of 24:
let n represent the number of slips of paper (each containing a 4) that we have to add to the original 10.
The sum then becomes: 24 + 4n
The number of slips of paper becomes 10 + n.
The average becomes the sum divided by the number of slips of paper: (24 + 4n) / (10 + n).
Since we want the average to be 3, we can create this equation: (24 + 4n) / (10 + n) = 3
Solving this by multiplying both sides by (10 + n) ---> 24 + 4n = 3(10 + n)
---> 24 + 4n = 30 + 3n
---> n = 6
We would have to add 6 slips to the original 10.
How many 4s must be added to get an average of at least 3.5.
Using the technique of the previous problem, we get the equation: (24 + 4n) / (10 + n) > 3.5
Can you take it from here?