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# Super HARD expected value questions!

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Deleted - go an abuse someone elses site - Melody.

2. Suppose I have a bag with 12 slips of paper in it. Some of the slips have a 2 on them, and the rest have a 7 on them. If the expected value of the number shown on a slip randomly drawn from the bag is 3.25, then how many slips have a 2?

3. I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different balls at random. What is the expected value of the sum of the numbers on the marbles?

4. Two standard six-faced dice are rolled. Cara scores x points if the sum of the numbers rolled is greater than or equal to their product, otherwise Jeremy scores one point. What should be the value of x to make the game fair?

5. An art starts walking from the origin. During each second, it has an equal chance of moving 1 inch right, 2 inches up, 3 inches left, or 4 inches down. What is the ant's expected location after 5 seconds if it starts at the origin?

6. A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 1 through 4. If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins \$1. If the number chosen appears on the bottom of both of the dice, then the player wins \$2. If the number chosen does not appear on the bottom of either of the dice, the player loses \$1.  What is the expected return to the player, in dollars, for one roll of the dice?

deleted

May 22, 2019
edited by Melody  May 22, 2019

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2. Suppose I have a bag with 12 slips of paper in it. Some of the slips have a 2 on them, and the rest have a 7 on them. If the expected value of the number shown on a slip randomly drawn from the bag is 3.25, then how many slips have a 2?

Let  N be the number that have a 2  on them.....then 12- N have a (7)

So we have that

(2)(N/12)  + (7)(12 - N)/12  =  3.25     multiply  through by 12

2N + 7(12 - N)  =  39

2N  + 84  - 7N  = 39      subtract 84 from both sides

-5N  = -45        divide both sides by -5

N  = 9  =  the number of slips with "2" on them   May 22, 2019
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3. I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different balls at random. What is the expected value of the sum of the numbers on the marbles?

We have  P(5,2)  = 20 possible ways to select the marbles

(1,2)  (2,3)  (3,4)  (4,5)

(2,1)  (3,2)  (4,3)   (5,4)

(1,3)  (2,4)  (3,5)

(3,1)  (4,2)  (5,3)

(1,4)  (2,5)

(4,1)  (5,2)

(1,5)

(5,1)

So the expected value of the sum is

3(2/20) + 4(2/20) + 5(4/20) + 6(4/20) + 7(4/20) + 8(2/20) + 9(2/20)  =  6

CORRECTED ANSWER....THANKS TO THE GUEST FOR POINTING OUT MY LOGIC FLAW !!!!   May 22, 2019
edited by CPhill  May 24, 2019
edited by CPhill  May 24, 2019
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Cphill I don't think you can pick 2 twice, you have to pick two different balls. Your sum can never be 2 because to get 2 you have to pick ball number 1 twice.

Guest May 24, 2019
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Yuck!!!....definitely didn't think on that one!!!!!

Let me re-work it  !!!!   CPhill  May 24, 2019
edited by CPhill  May 24, 2019
edited by CPhill  May 24, 2019
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5. An art starts walking from the origin. During each second, it has an equal chance of moving 1 inch right, 2 inches up, 3 inches left, or 4 inches down. What is the ant's expected location after 5 seconds if it starts at the origin?

[  1(.25) - 3(.25) , 2(.25) - 4(.25) ]   =  [ -.5, -.5 ]

So....each second, the ant  is expected to move 1/2 unit to the left and 1/2 unit down

So....after 5 seconds  the expected location is  5 [ -.5, -.5 ]  =  (-2.5, -2.5)   May 22, 2019
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Jerryjing2006   -   You are a lazy cheating child who wants all your homework done for you.

Why do you persist in fascilitating cheating Chris.

May 22, 2019
edited by Melody  May 22, 2019
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I'm sorry, but I think you've misunderstood me. These are my friends HW answers and he has asked me for his help. I have already completed all of his HW to the best of my ability. But I've posted these questions to further my knowledge on these topics and also help out a friend. I hope you understand!

jerryjing2006  May 22, 2019
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Yes I understand perfectly well.

You have posted no evidence that you have attempted any of these yourself.

The evidence clearly suggests that you are a cheat and a liar.

Melody  May 22, 2019
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4. Two standard six-faced dice are rolled. Cara scores x points if the sum of the numbers rolled is greater than or equal to their product, otherwise Jeremy scores one point. What should be the value of x to make the game fair?

Here is a table   comparing  sums and products

1     2     3     4    5    6

1    G    G    G    G   G   G

2    G    E    L     L    L    L

3    G    L    L     L     L   L

4    G    L    L     L    L    L

5    G   L     L     L    L    L

6    G   L     L     L    L    L

Cara has a (12/36)  = (1/3) chance of having a greater or equal sum to a respective product

Jeremy has a (2/3) chance  = twice Cara's chance

So

1 (2/3)   =  x (1/3)

(2/3)  = (1/3) x       divide both sides by (1/3)

(2/3) / (1/3)  =  x

(2/1)  =  x  =  2 ⇒   Cara should get 2 points for a winning turn to make the game fair   May 22, 2019
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6. A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered 1 through 4. If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins 1 dollar. If the number chosen appears on the bottom of both of the dice, then the player wins 2 dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses 1 dollar. What is the expected return to the player, in dollars, for one roll of the dice? --7H3_5H4D0W

May 22, 2019
edited by GAMEMASTERX40  May 22, 2019
edited by GAMEMASTERX40  May 22, 2019