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 $11$?
(a) The probability of drawing a $5$ is $\frac{8}{10}$ and the probability of drawing a $12$ is $\frac{2}{10}$. Therefore, the expected value is $E = \frac{8}{10} \cdot 5 + \frac{2}{10} \cdot 12 = 4 + \frac{24}{10} = 4 + \frac{12}{5} = \frac{32}{5} = 6.4$. So, the expected value of the number shown when we draw a single slip of paper is $6.4$.
(b) If we add one additional $12$ to the bag, then there will be a total of $3$ slips with a $12$ and $8$ slips with a $5$. Now, the expected value is $E = \frac{8}{11} \cdot 5 + \frac{3}{11} \cdot 12 = \frac{40}{11} + \frac{36}{11} = \frac{76}{11} = 6.909$. So, the expected value of the number shown with an additional $12$ is $6.909$.
(c) If we add two additional $12$'s to the bag, then there will be a total of $4$ slips with a $12$ and $8$ slips with a $5$. Now, the expected value is $E = \frac{8}{12} \cdot 5 + \frac{4}{12} \cdot 12 = \frac{40}{12} + \frac{48}{12} = \frac{88}{12} = 7.333$. So, the expected value of the number shown with two additional $12$'s is $7.333$.
(d) Let $x$ be the number of $12$'s added to make the expected value equal to $10$. We have the equation $\frac{8}{10 + x} \cdot 5 + \frac{x}{10 + x} \cdot 12 = 10$. Solving for $x$ gives us $\frac{40}{10 + x} + \frac{12x}{10 + x} = 10$. Multiplying both sides by $10 + x$, we get $40 + 12x = 10(10 + x)$. Expanding and simplifying, we have $40 + 12x = 100 + 10x$. Subtracting $10x$ from both sides, we get $2x = 60$, so $x = 30$. Therefore, we need to add $30$ $12$'s to make the expected value equal to $10$.
(e) Let $x$ be the number of $12$'s added before the expected value is at least $11$. We have the inequality $\frac{8}{10+x} \cdot 5 + \frac{x}{10+x} \cdot 12 \geq 11$. Multiplying both sides by $10 + x$, we get $40 + 12x \geq 11(10 + x)$. Expanding and simplifying, we have $40 + 12x \geq 110 + 11x$. Subtracting $11x$ from both sides, we get $x \geq 70$. Therefore, we need to add at least $70$ $12$'s before the expected value is at least $11$.
To find the expected value, we multiply each outcome by its corresponding probability and sum all of them. Let's go through each part of the question step by step.
(a) In the original scenario, we have 8 slips with a 5 and 2 slips with a 12 in the bag. The probabilities of drawing a slip with a 5 and a 12 are $8/10$ and $2/10$, respectively. So, the expected value is:
$E = \left(\frac{8}{10} \cdot 5\right) + \left(\frac{2}{10} \cdot 12\right) = 4 + 2.4 = 6.4$
(b) If we add one more slip with a 12 to the bag, the total number of slips becomes 11. Now we have 8 slips with a 5 and 3 slips with a 12. The new probabilities are $8/11$ for 5 and $3/11$ for 12. The expected value becomes:
$E = \left(\frac{8}{11} \cdot 5\right) + \left(\frac{3}{11} \cdot 12\right) = 3.636... + 3.272... = 6.909...$
(c) If we add two more slips with a 12 to the bag, the total number of slips becomes 12. Now we have 8 slips with a 5 and 4 slips with a 12. The new probabilities are $8/12$ for 5 and $4/12$ for 12. The expected value becomes:
$E = \left(\frac{8}{12} \cdot 5\right) + \left(\frac{4}{12} \cdot 12\right) = 3.333... + 4 = 7.333...$
(d) To make the expected value equal to 10, we need to find how many slips with a 12 we should add. The original expected value is 6.4, and each additional slip with a 12 increases the expected value by 6. So, we can set up an equation:
$6.4 + 6k = 10$
$6k = 10 - 6.4$
$6k = 3.6$
$k = \frac{3.6}{6} = 0.6$
Since we can't have a fraction of a slip, we need to add at least one slip with a 12 to make the expected value 10.
(e) To find how many slips with a 12 we need to add before the expected value is at least 11, we can set up a similar equation. Let's call the number of slips we need to add x:
$6.4 + 6x = 11$
$6x = 11 - 6.4$
$6x = 4.6$
$x = \frac{4.6}{6} = 0.766...$
Again, since we can't have a fraction of a slip, we need to add at least one slip with a 12 to make the expected value at least 11.
(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 $n=\boxed{8}$.