Question

marlery 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 losing

There are 4 non - blue marbles out of 5 total marbles. So the probability of losing $P(\text{lose})=\frac{4}{5}$. And the point value for losing is $- 2$ points.

Step2: Calculate probability of winning

The probability of winning $P(\text{win})=\frac{1}{5}$. Let the point value for winning be $x$ points.

Step3: Set up expected - value equation for a fair game

For a fair game, the expected value $E(X)=0$. The formula for expected value is $E(X)=P(\text{win})\times x+P(\text{lose})\times(-2)$. Substituting the probabilities, we get $0=\frac{1}{5}x+\frac{4}{5}\times(-2)$.

Step4: Solve the equation for $x$

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

Answer:

C. 8