Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them.
How many $9$'s do I have to add before the expected value is at least $8$?
The answer isn't 10. It is 80 - 24 - 18 = 38.
Is this copied from AoPS? I can see the exact same question there. If you need help you shouldn't come here.
I think that is what Damon meant. I think Damon was just wondering what $10$ meant.
Bruh this is AOPS
To solve this problem, we need to find the expected value of a random variable representing the value of a slip of paper drawn from the bag. The expected value is calculated by multiplying each possible value of the random variable by its probability, and then summing these products.
Let's denote the number of additional $9$'s we need to add as $x$. After adding $x$ $9$'s, there will be a total of $10 + x$ slips of paper in the bag, with $8$ slips having a $3$ and $2 + x$ slips having a $9$.
The probability of drawing a $3$ after adding $x$ $9$'s is $\frac{8}{10 + x}$. Similarly, the probability of drawing a $9$ is $\frac{2 + x}{10 + x}$.
The expected value is then:
\[E = \frac{8}{10 + x} \cdot 3 + \frac{2 + x}{10 + x} \cdot 9.\]
We want to find the value of $x$ for which $E \geq 8$. So we solve the inequality:
\[E \geq 8.\]
Substituting the values, we have:
\[\frac{8}{10 + x} \cdot 3 + \frac{2 + x}{10 + x} \cdot 9 \geq 8.\]
Multiplying both sides by $10 + x$, we get:
\[24 + 9(2 + x) \geq 8(10 + x).\]
Simplifying, we have:
\[24 + 18 + 9x \geq 80 + 8x.\]
Combining like terms, we get:
\[42 + 9x \geq 80 + 8x.\]
Subtracting $8x$ from both sides, we have:
\[42 + x \geq 80.\]
Subtracting $42$ from both sides, we get:
\[x \geq 38.\]
Therefore, you will need to add at least $38$ $9$'s before the expected value is at least $8$.
$10$ slips of paper
How many ???????
Oh , it must be 10
8 = [8*3 + 9*2 + 9*n ] /(10 + n)
80 + 8 n = 24 + 18 + 9 n
n = 80 - 24 - 18