Question

finding the expected value in a marble game
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 is the expected value of the points for derek and mia?
p(gy) =
completed
done

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
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
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: $-\frac{1}{2}$; $P(GY)=\frac{1}{4}$