finding the expected value in a marble game\nyellow marble in a bag\nsomebody picks a marble out of the bag…
finding the expected value in a marble game\nyellow marble in a bag\nsomebody 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 is the expected value of the points for derek and mia?\np(gy) = \ncompleted\ndone
Answer
Explanation:
Step1: Calculate total number of outcomes
There are 2 colors of marbles. When picking 2 marbles with replacement, the total number of outcomes is $2\times2 = 4$ (GG, GY, YG, YY).
Step2: Determine favorable outcomes for P(GY)
The event of getting a green - then - yellow marble (GY) is 1 out of 4 possible outcomes. So $P(GY)=\frac{1}{4}$.
Step3: Set up probability distribution for points
Let $X$ be the number of points for Derek.
| Outcome | Probability | Derek's Points |
|---|---|---|
| GG | $\frac{1}{4}$ | 1 |
| GY | $\frac{1}{4}$ | 1 |
| YG | $\frac{1}{4}$ | 1 |
| YY | $\frac{1}{4}$ | - 1 |
Step4: Calculate Derek's expected value
The expected value formula is $E(X)=\sum_{i}x_ip_i$. $E(X)=1\times\frac{1}{4}+1\times\frac{1}{4}+1\times\frac{1}{4}+(-1)\times\frac{1}{4}=\frac{1 + 1+1 - 1}{4}=\frac{2}{4}=\frac{1}{2}$. For Mia, her points are the opposite of Derek's in each case. Let $Y$ be the number of points for Mia.
| Outcome | Probability | Mia's Points |
|---|---|---|
| GG | $\frac{1}{4}$ | - 1 |
| GY | $\frac{1}{4}$ | - 1 |
| YG | $\frac{1}{4}$ | - 1 |
| YY | $\frac{1}{4}$ | 1 |
| $E(Y)=(-1)\times\frac{1}{4}+(-1)\times\frac{1}{4}+(-1)\times\frac{1}{4}+1\times\frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}$. |
Answer:
Derek's expected value: $\frac{1}{2}$; Mia's expected value: $-\frac{1}{2}$; $P(GY)=\frac{1}{4}$