(a) Since there are $10$ slips of paper in the bag, the probability of drawing each slip is $\frac{1}{10}$. The expected value is then \[
\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 5\right)+\left(\frac{1}{10} \cdot 12\right)+\left(\frac{1}{10} \cdot 12\right)=\frac{8 \cdot 5+2 \cdot 12}{10}=\boxed{6}.\] (b) If we add one additional $12$ to the bag, then there are $11$ slips of paper total. The expected value is then \[
\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 5\right)+\left(\frac{1}{11} \cdot 12\right)+\left(\frac{1}{11} \cdot 12\right)+\left(\frac{1}{11} \cdot 12\right)=\frac{8 \cdot 5+3 \cdot 12}{11}=\boxed{\frac{15}{11}}.\] (c) If we add two additional $12$'s to the bag, then there are $12$ slips of paper total. The expected value is then \[
\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 5\right)+\left(\frac{1}{12} \cdot 12\right)+\left(\frac{1}{12} \cdot 12\right)+\left(\frac{1}{12} \cdot 12\right)+\left(\frac{1}{12} \cdot 12\right)=\frac{8 \cdot 5+4 \cdot 12}{12}=\boxed{\frac{11}{3}}.\] (d) Let $n$ be the number of $12$'s we need to add to the bag. There are then $10+n$ slips of paper total. Setting the expected value equal to $10$, we have \[
\frac{8 \cdot 5+n \cdot 12}{10+n}=10.
\] Multiplying both sides by $10+n$, we get $8 \cdot 5+n \cdot 12=100+10n$. Subtracting $8 \cdot 5+100$ from both sides and factoring the left side gives $4n=20$, or $n=\boxed{5}$. Note that our found answer follows from the observation that the expected value of the number shown increases by $1$ for each $12$ we add. (e) Let $n$ be the number of $12$'s we need to add to the bag. There are then $10+n$ slips of paper total. Setting the expected value equal to $11$, we have \[
\frac{8 \cdot 5+n \cdot 12}{10+n}=11.
\] We can solve this equation in the same manner we solved equation (d) to get $n=\boxed{8}$.
Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $5$ on them and the other two have an $12$ on them.
(a) What is the expected value of the number shown when we draw a single slip of paper?
(b) What is the expected value of the number shown if we add one additional $12$ to the bag?
(c) What is the expected value of the number shown if we add two additional $12$'s (instead of just one) to the bag?
(d) How many $12$'s do we have to add to make the expected value equal to $10$?
(e) How many $12$'s do we have to add before the expected value is at least $11$?
1 answer