1. 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.
What is the expected value of the number shown when we draw a single slip of paper?
2. 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.
What is the expected value of the number shown if we add one additional $9$ to the bag?
3. 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.
What is the expected value of the number shown if we add two additional $9$'s (instead of just one) to the bag?
4. 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 we have to add to make the expected value equal to $6$?
Thanks!
For number 2:
Question: 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. What is the expected value of the number shown if we add one additional $9$ to the bag?
Answer: 3(8/11) + 9 (3/11) = 24/11 + 27 /11 = 51/11 ≈ 4.63
For number 3:
Question: 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. What is the expected value of the number shown if we add two additional 9's (instead of just one) to the bag?
Answer: 3 ( 8/12) + 9(4/12) = 24/12 + 36/12 = 60/12 = 5
For number 1:
Question: 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. What is the expected value of the number shown when we draw a single slip of paper?
Answer: 3(8/10) + 9(2/10) = 24/10 + 18/10 = 42/10 = 21/5 = 4.2
For number 4:
Question: 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 we have to add to make the expected value equal to 6?
Answer: The expected value being 6 would imply that the amounts of 3 and 9 must be equal to make the probabilities be equal, which is done by adding 6 9’s. So the answer is 6.
To find the expected value, we need to calculate the probability of each outcome and multiply it by the corresponding value. Let's solve each question step by step:
1. The bag has 10 slips of paper, out of which 8 have a 3 and 2 have a 9. To find the expected value, we need to calculate the weighted average of the possible outcomes. The probability of drawing a slip with a 3 is 8/10, and the probability of drawing a slip with a 9 is 2/10.
Expected value = (Probability of getting a 3 * Value of 3) + (Probability of getting a 9 * Value of 9)
= (8/10 * 3) + (2/10 * 9)
= 24/10 + 18/10
= 42/10
= 4.2
Therefore, the expected value of the number shown when drawing a single slip from the bag is 4.2.
2. If we add one additional 9 to the bag, the total number of slips will be 11. The probability of drawing a slip with a 3 remains the same: 8/11. The probability of drawing a slip with a 9 is now 3/11 (2 original 9's + 1 additional 9 / total 11 slips).
Expected value = (Probability of getting a 3 * Value of 3) + (Probability of getting a 9 * Value of 9)
= (8/11 * 3) + (3/11 * 9)
= 24/11 + 27/11
= 51/11
= 4.636363
Therefore, the expected value of the number shown if we add one additional 9 to the bag is approximately 4.636363.
3. If we add two additional 9's to the bag, the total number of slips will be 12. The probability of drawing a slip with a 3 remains 8/12, and the probability of drawing a slip with a 9 is now 4/12 (2 original 9's + 2 additional 9's / total 12 slips).
Expected value = (Probability of getting a 3 * Value of 3) + (Probability of getting a 9 * Value of 9)
= (8/12 * 3) + (4/12 * 9)
= 24/12 + 36/12
= 60/12
= 5
Therefore, the expected value of the number shown if we add two additional 9's to the bag is 5.
4. To make the expected value equal to 6, we need to find out how many additional 9's to add.
Expected value = (Probability of getting a 3 * Value of 3) + (Probability of getting a 9 * Value of 9)
Let x be the number of additional 9's.
Expected value = (8/10 * 3) + ((2 + x)/ (10 + x) * 9)
Setting this equal to 6 and solving for x:
(8/10 * 3) + ((2 + x)/ (10 + x) * 9) = 6
Multiplying through by (10 + x) to remove the denominator:
(8/10 * 3) * (10 + x) + ((2 + x) * 9) = 6 * (10 + x)
Simplifying and solving for x:
24 + 2.4x + 18 + 9x = 60 + 6x
11.4x + 42 = 60 + 6x
11.4x - 6x = 60 - 42
5.4x = 18
x = 18/5.4
x = 3.333333
Therefore, we need to add approximately 3.333333 (or 4) 9's to make the expected value equal to 6.
Note: Since we cannot have fractional slips, we round up the result. This means that adding 4 9's will give an expected value slightly higher than 6, but it will be the closest whole number value that satisfies the condition.
the reason for the dollar signs is because they copy pasted from AoPS which uses Latex. STOP CHEATING ON YOUR HOMEWORK!!!
It would help if you proofread your questions before you posted them.
First of all, the dollar sign only goes BEFORE the value, e.g., $10.
Second, if it is not a dollar amount, do not use a dollar sign ($).