Let's solve each problem step by step from the image you uploaded. The problems are about
expected value, which is a fundamental concept in probability.
---
Problem 1:
> The expected payout for each play of a carnival game is $0.42. If each game costs $0.50 to play, what is the expected net gain (or loss) per play?
####
Solution:
- Expected payout = $0.42
- Cost to play = $0.50
The
net gain is:
$$
\text{Expected net gain} = \text{Expected payout} - \text{Cost to play}
= 0.42 - 0.50 = -0.08
$$
So, the
expected net loss is
$0.08 per play.
✔ Answer: $-0.08 (a loss of 8 cents per play)
---
Problem 2:
> A tile is drawn at random from a bag and replaced. The probability of drawing a blue tile is 0.60 and the probability of drawing a red tile is 0.20. What is the expected number of blue tiles in 10 draws?
####
Solution:
We're told:
- P(blue) = 0.60
- P(red) = 0.20
- But there might be other colors? Let’s check:
Since probabilities must sum to 1:
$$
P(\text{other}) = 1 - 0.60 - 0.20 = 0.20
$$
But we’re only asked for the
expected number of blue tiles in 10 draws.
Each draw is independent, and replacement is used.
Expected number of blue tiles in one draw = 0.60
So in 10 draws:
$$
E = 10 \times 0.60 = 6
$$
✔ Answer: 6 blue tiles
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Problem 3:
> An individual die has the numbers 0 through 9 on its faces. What is the expected value when rolling a ten-sided die? Is this expected value a possible outcome?
####
Solution:
A fair 10-sided die with faces numbered 0 through 9.
All outcomes are equally likely.
Expected value (E) is:
$$
E = \frac{0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{10}
= \frac{45}{10} = 4.5
$$
Now, is 4.5 a possible outcome? No — the die only shows integers from 0 to 9.
✔ Answer:
- Expected value =
4.5
- Is it a possible outcome?
No
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Problem 4:
> A bag has 6 red marbles, 3 blue marbles, and 1 orange marble. It's a game to randomly select a marble. You win $2 if you pick a red marble, $3 if you pick a blue marble, and $5 if you pick an orange marble. What is the expected value of the game?
####
Solution:
Total marbles = 6 (red) + 3 (blue) + 1 (orange) =
10
Probabilities:
- P(red) = 6/10 = 0.6 → Win $2
- P(blue) = 3/10 = 0.3 → Win $3
- P(orange) = 1/10 = 0.1 → Win $5
Expected value:
$$
E = (0.6)(2) + (0.3)(3) + (0.1)(5)
= 1.2 + 0.9 + 0.5 = 2.6
$$
✔ Answer: Expected value =
$2.60
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✔ Final Answers Summary:
1.
Expected net gain/loss: $-0.08
2.
Expected number of blue tiles in 10 draws: 6
3.
Expected value of die roll: 4.5;
Not a possible outcome
4.
Expected value of marble game: $2.60
Let me know if you want these written neatly or explained further!
Parent Tip: Review the logic above to help your child master the concept of expected value worksheet.
