Question

select all the correct answers.
which of the following situations are fair?

jack and mia both want the last cookie and neither will agree to share by splitting it. they ask a stranger passing by to flip a coin to decide who gets the cookie.

jericho designs a game for a school carnival. the game consists of a box of 50 colored balls; 35 are violet, 10 are orange, and 5 are yellow. the player has to choose one ball from the box. the player wins $5 if it is a yellow ball, otherwise, the player wins nothing.

ryan designs a game where a prize wheel is split into five equal sections. four sections are red and one is green. if the wheel is spun and lands on a red section, the players loses $1. if the wheel lands on green section, the player wins $5.

to decide which citizens will be asked to participate in a county - wide poll, each citizen is assigned a number and the county uses a random number generator to determine the citizens who will be contacted.

jason plays a game in which he has to pick a ball from a box of 10 balls, which contains 7 black balls and 3 white balls. he wins the game if he draws a white ball in one attempt.

ray is playing a game in which he rolls a six - sided number cube. if the outcome is six, he is paid $5. otherwise, he loses $1.

Explanation:

To determine if a situation is fair, we check if each outcome has an equal probability of occurring or if the expected value (for games involving winning/losing) is balanced.

Step 1: Analyze Jack and Mia's situation

A coin flip has two outcomes (heads/tails) with equal probability ($\frac{1}{2}$ each). So this is fair.

Step 2: Analyze Jericho's game

Total balls = 50. Probability of yellow (win) = $\frac{5}{50} = 0.1$, probability of non - yellow (lose) = $\frac{45}{50}=0.9$. The player wins $5$ with low probability and loses nothing with high probability. Not fair.

Step 3: Analyze Ryan's game

Prize wheel has 5 sections: 4 red, 1 green. Probability of red (lose $1$) = $\frac{4}{5}$, probability of green (win $5$) = $\frac{1}{5}$. Expected value for player: $(\frac{4}{5})(- 1)+(\frac{1}{5})(5)=\frac{-4 + 5}{5}=\frac{1}{5}>0$. But for fairness in a game (if it's a zero - sum or balanced), let's check the expected loss/gain. However, the key is the probability of winning and losing. The probability of losing is higher, but the expected value is positive. Wait, actually, for a fair game (in terms of expected value), the expected value should be zero. But maybe the question is about equal chance of winning/losing. Wait, no, in Ryan's game, the probability of red (losing $1$) is $\frac{4}{5}$ and green (winning $5$) is $\frac{1}{5}$. The expected value is $(\frac{4}{5})(-1)+(\frac{1}{5})(5)=\frac{- 4 + 5}{5}=\frac{1}{5}$. But maybe the question is about equal probability of success and failure. Wait, no, let's re - evaluate.

Wait, the first situation: coin flip, equal probability. The fourth situation: random number generator, each citizen has an equal chance of being selected (since it's random). Let's check the sixth situation: Ray's game. A six - sided die. Probability of 6 (win $5$) is $\frac{1}{6}$, probability of non - 6 (lose $1$) is $\frac{5}{6}$. Expected value: $(\frac{1}{6})(5)+(\frac{5}{6})(-1)=\frac{5 - 5}{6}=0$. So this is a fair game (expected value is zero).

Wait, let's re - check each:

1. Jack and Mia: Coin flip, equal probability ($\frac{1}{2}$ each) → fair.
2. Jericho's game: As above, not fair.
3. Ryan's game: As above, expected value is $\frac{1}{5}$, but the probabilities of winning and losing are not equal. Wait, maybe I made a mistake. Let's recalculate Ryan's expected value: $E = P(red)\times(-1)+P(green)\times(5)=\frac{4}{5}\times(-1)+\frac{1}{5}\times5=\frac{-4 + 5}{5}=\frac{1}{5}$. But is this a fair game? If we consider a fair game as one where the expected value for both sides (if it's a two - player game) is zero, but here it's a player vs the game. Maybe the question is about equal chance of winning (success) and losing (failure). In Ryan's game, chance of winning (green) is $\frac{1}{5}$, chance of losing (red) is $\frac{4}{5}$ → not equal. So not fair.
4. County poll: Random number generator gives each citizen an equal chance of being selected → fair.
5. Jason's game: Total balls = 10, 7 black, 3 white. Probability of white (win) = $\frac{3}{10}$, black (lose) = $\frac{7}{10}$ → not equal → not fair.
6. Ray's game: Probability of 6 (win $5$) = $\frac{1}{6}$, probability of non - 6 (lose $1$) = $\frac{5}{6}$. Expected value: $(\frac{1}{6})(5)+(\frac{5}{6})(-1)=\frac{5 - 5}{6}=0$. So this is a fair game (expected value is zero).

Wait, maybe I missed some. Let's re - check:

- Jack and Mia: Fair (equal probability).
- Jericho: Not fair (low probability of winning).
- Ryan: Let's see, the payout for green is $5$ and for red is - $1$. The expected value for the player is $\frac{4}{5}\times(-1)+\frac{1}{5}\times5=\frac{-4 + 5}{5}=0.2$. But maybe the question is about equal chance of winning (success) and losing (failure). No, in Ryan's case, chance of winning is $\frac{1}{5}$, losing is $\frac{4}{5}$.
- County poll: Fair (random selection, equal chance).
- Jason: Probability of winning is $\frac{3}{10}$, losing is $\frac{7}{10}$ → not fair.
- Ray: Expected value is zero. Let's calculate again: $E=\frac{1}{6}\times5+\frac{5}{6}\times(-1)=\frac{5 - 5}{6}=0$. So this is fair.

Wait, maybe the first, fourth, and sixth are fair? Wait, no, let's check the first: coin flip, equal probability for Jack and Mia to get the cookie → fair. Fourth: random number generator, each citizen has an equal chance of being chosen → fair. Sixth: expected value is zero → fair. Wait, but let's check the sixth again. The die has 6 sides. If we consider the game as a bet, the expected value is zero, so it's a fair bet.

Wait, maybe I made a mistake with Ryan's game. Let's see, the wheel has 5 sections. 4 red (lose $1$) and 1 green (win $5$). The expected value for the player is $(\frac{4}{5})(-1)+(\frac{1}{5})(5)=\frac{-4 + 5}{5}=0.2$. So the player has a positive expected value, but is the game fair? If the game is between the player and the house, a fair game would have expected value zero. So Ryan's game is not fair.

Now, let's list the fair ones:

1. Jack and Mia: Coin flip, equal probability.
2. County poll: Random selection, equal chance.
3. Ray's game: Expected value zero.

Wait, but let's check Ray's game again. The probability of rolling a six is $\frac{1}{6}$, and the probability of not rolling a six is $\frac{5}{6}$. The payout for six is $5$, and for non - six is - $1$. The expected value is $(\frac{1}{6})\times5+(\frac{5}{6})\times(-1)=\frac{5 - 5}{6}=0$. So it's a fair game.

Now, let's check Jason's game: 10 balls, 7 black, 3 white. Probability of white (win) is $\frac{3}{10}$, black (lose) is $\frac{7}{10}$. Not equal, so not fair.

Jericho's game: 5 yellow out of 50. Probability of winning is $\frac{5}{50}=0.1$, losing is $0.9$. Not fair.

Ryan's game: As above, not fair.

So the fair situations are:

- Jack and Mia's coin flip.
- County poll (random number generator).
- Ray's die game.

Wait, but let's confirm the county poll. Using a random number generator to select citizens means each citizen has an equal probability of being selected, so it's a fair way to choose participants.

Now, let's check the options:

1. Jack and Mia: Fair.
2. Jericho: Not.
3. Ryan: Not.
4. County poll: Fair.
5. Jason: Not.
6. Ray: Fair.

So the correct answers are the first, fourth, and sixth options.

Answer:

• Jack and Mia both want the last cookie and neither will agree to share by splitting it. They ask a stranger passing by to flip a coin to decide who gets the cookie.
• To decide which citizens will be asked to participate in a county - wide poll, each citizen is assigned a number and the county uses a random number generator to determine the citizens who will be contacted.
• Ray is playing a game in which he rolls a six - sided number cube. If the outcome is six, he is paid $5. Otherwise, he loses $1.