Question

finding the expected value in a marble game
derek and mia place two green marbles and one yellow marble in a bag. somebody picks a marble out of the bag without looking and records its color (g for green and y for yellow). they replace the marble and then pick another marble. if the two marbles picked have the same color, derek loses 1 point and mia gains 1 point. if they are different colors, mia loses 1 point and derek gains 1 point. what

Explanation:

Step1: Calculate total number of outcomes

There are 3 marbles initially. For the first - pick, there are 3 choices, and for the second - pick, there are also 3 choices. So the total number of outcomes is $n = 3\times3=9$.

Step2: Calculate number of same - color outcomes

Green - green: The probability of picking a green on the first pick is $\frac{2}{3}$, and the probability of picking a green on the second pick is also $\frac{2}{3}$. So the number of green - green outcomes is $2\times2 = 4$. Yellow - yellow: The probability of picking a yellow on the first pick is $\frac{1}{3}$, and the probability of picking a yellow on the second pick is also $\frac{1}{3}$. So the number of yellow - yellow outcomes is $1\times1 = 1$. The total number of same - color outcomes is $4 + 1=5$.

Step3: Calculate number of different - color outcomes

The number of different - color outcomes is $9-5 = 4$.

Step4: Define the value of each outcome

Let $X$ be the value of Ma's score. If the colors are the same, Ma gains 1 point ($X = 1$), and the probability $P(X = 1)=\frac{5}{9}$. If the colors are different, Ma loses 1 point ($X=-1$), and the probability $P(X=-1)=\frac{4}{9}$.

Step5: Calculate the expected value

The formula for the expected value $E(X)$ is $E(X)=\sum_{i}x_{i}P(x_{i})$. So $E(X)=1\times\frac{5}{9}+(-1)\times\frac{4}{9}=\frac{5 - 4}{9}=\frac{1}{9}$.

Answer:

$\frac{1}{9}$