Question

Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $7$ on them and the other two have a $2$ on them.

What is the smallest number of $2$'s we could add to the bag to make expected value of a slip from the bag less than $5$?

Answer 1

Let x be the number of additional 2's. The new expected value is
(8·7 + (2+x)·2) / (10 + x) = (60 + 2x)/(10 + x).

Require (60 + 2x)/(10 + x) < 5:
60 + 2x < 50 + 5x ⇒ 10 < 3x ⇒ x > 10/3.

The smallest integer x satisfying this is 4. Thus add 4 twos.