At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

As a project for math class, two students devised a game in which 3 black marbles and 2 red marbles are put into a bag. First, the players must decide who is playing black marbles and who is playing red marbles. Then each player takes a turn at drawing a marble, noting the color, replacing the marble in the bag, and then drawing a second marble and noting the color before returning it to the bag. The point scheme for the game is detailed in the table below.

\begin{tabular}{|l|l|}
\hline \multicolumn{2}{|c|}{Point Values for Marble Game} \\
\hline \multicolumn{1}{|c|}{Black Marble Points} & \multicolumn{1}{c|}{Red Marble Points} \\
\hline Both black: +2 points & Both red: +4 points \\
\hline Different colors: -1 point & Different colors: -1 point \\
\hline Both red: 0 points & Both black: 0 points \\
\hline
\end{tabular}

If Seth is challenged to a game by a classmate, which statement below is correct in all aspects to help him make the correct choice?

A. Since [tex]E(\text{black}) = 0.24[/tex] and [tex]E(\text{red}) = 0.16[/tex], Seth should choose to play black marbles.

B. [tex]E(\text{red})[/tex] will be twice that of [tex]E(\text{black})[/tex], so Seth should choose to play red marbles.

C. Since [tex]E(\text{red}) = E(\text{black})[/tex], it is a fair game, so it doesn't matter which color Seth chooses.

D. Both options will lose points because there are two ways to lose points and only one way to gain points.

Sagot :

Let's analyze the given problem step-by-step in a structured manner by calculating the expected values (E) for choosing black marbles and red marbles.

### Defining Probabilities
First, we define the probabilities of drawing each color of marble:
- Probability of drawing a black marble ([tex]\( P(\text{black}) \)[/tex]) is [tex]\(\frac{3}{5}\)[/tex].
- Probability of drawing a red marble ([tex]\( P(\text{red}) \)[/tex]) is [tex]\(\frac{2}{5}\)[/tex].

### Calculating the Expected Value for Black Marble Player
The expected value (E) for the player choosing black marbles considers all possible outcomes:
1. Drawing two black marbles (Both black): [tex]\( P(\text{black}) \times P(\text{black}) \)[/tex]
2. Drawing one black and one red marble (Different colors): [tex]\( 2 \times P(\text{black}) \times P(\text{red}) \)[/tex] (two ways)

#### Points Assigned to Outcomes for Black Marbles
- If both drawn marbles are black (Black, Black): +2 points
- If one marble is black and the other is red (Black, Red or Red, Black): -1 point
- If both are red (Red, Red): 0 points

#### Calculation:
[tex]\[ E(\text{black}) = (P(\text{black} \cap \text{black}) \times 2) + (P(\text{black} \cap \text{red}) \times -1) + (P(\text{red} \cap \text{red}) \times 0) \][/tex]
[tex]\[ E(\text{black}) = \left(\frac{3}{5} \cdot \frac{3}{5} \times 2\right) + \left(2 \cdot \frac{3}{5} \cdot \frac{2}{5} \times -1\right) + \left(\frac{2}{5} \cdot \frac{2}{5} \times 0\right) \][/tex]

### Calculating the Expected Value for Red Marble Player
The expected value (E) for the player choosing red marbles also considers all possible outcomes:
1. Drawing two red marbles (Both red): [tex]\( P(\text{red}) \times P(\text{red}) \)[/tex]
2. Drawing one red and one black marble (Different colors): [tex]\( 2 \times P(\text{red}) \times P(\text{black}) \)[/tex] (two ways)

#### Points Assigned to Outcomes for Red Marbles
- If both drawn marbles are red (Red, Red): +4 points
- If one marble is red and the other is black (Red, Black or Black, Red): -1 point
- If both are black (Black, Black): 0 points

#### Calculation:
[tex]\[ E(\text{red}) = (P(\text{red} \cap \text{red}) \times 4) + (P(\text{red} \cap \text{black}) \times -1) + (P(\text{black} \cap \text{black}) \times 0) \][/tex]
[tex]\[ E(\text{red}) = \left(\frac{2}{5} \cdot \frac{2}{5} \times 4\right) + \left(2 \cdot \frac{2}{5} \cdot \frac{3}{5} \times -1\right) + \left(\frac{3}{5} \cdot \frac{3}{5} \times 0\right) \][/tex]

After computing these values, we find:
- [tex]\( E(\text{black}) = 0.24 \)[/tex]
- [tex]\( E(\text{red}) = 0.16 \)[/tex]

### Conclusion
Since [tex]\( E(\text{black}) = 0.24 \)[/tex] and [tex]\( E(\text{red}) = 0.16 \)[/tex]:
- Seth should choose to play black marbles because the expected value for the person choosing black marbles is higher. Therefore, the correct statement helping Seth make the best choice is:

"Since [tex]\(E(\text{black}) = 0.24\)[/tex] and [tex]\(E(\text{red}) = 0.16\)[/tex], Seth should choose to play black marbles."