Question

marleny is creating a game of chance for her family. she has 5 different colored marbles in a bag: blue, red, yellow, white, and black. she decided that blue is the winning color. if a player chooses any other color, they lose 2 points. how many points should the blue marble be worth for the game to be fair?
4
6
8
10

Explanation:

Step1: Calculate probability of each outcome

The probability of drawing the blue marble is $P(\text{blue})=\frac{1}{5}$, and the probability of drawing a non - blue marble is $P(\text{non - blue})=\frac{4}{5}$.

Step2: Set up expected - value equation for a fair game

Let $x$ be the number of points the blue marble is worth. The expected value $E$ of the game should be 0 for it to be fair. The value of drawing a non - blue marble is $- 2$ points. So, $E = P(\text{blue})\times x+P(\text{non - blue})\times(-2)=0$. Substitute the probabilities: $\frac{1}{5}x+\frac{4}{5}\times(-2) = 0$.

Step3: Solve the equation for $x$

First, expand the equation: $\frac{1}{5}x-\frac{8}{5}=0$. Add $\frac{8}{5}$ to both sides: $\frac{1}{5}x=\frac{8}{5}$. Multiply both sides by 5 to get $x = 8$.

Answer:

8