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

Answer 1

The answer is four because if you lose two from the four you had you will still have a least 2 hope This helped

Answer 2

The points the blue marble should be worth for the game to be fair is given by: Option C: 8

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

How to calculate the expectation of a discrete random variable?

Expectation can be taken as a weighted mean, weights being the probability of occurrence of that specific observation.

Thus, if the random variable is X, and its probability mass function is given : f(x) = P(X = x), then we have:

$E(X) = \sum_{i=1}^n( f(x_i) \times x_i)$

(n is number of values X takes)

For the game to be fair, the expected number of points that one can loose should be equal to the expected number of points one can gain.

Now, probability of getting any specific colored marble = 1/5 = 0.2 (since there is one way to choose a specified colored marble, but there are 5 ways to choose a single (no color specified) marble out of 5 marbles.

The game would be fair if there is equal chance of winning and loosing.
Let we have:

• X = number of points one gets in this game,
• p = number of points that blue colored marble should have for the game to be fair.

Then we need E(X) = expectation of X = 0 (since the negative points of loosing would equate to positive point of winning, thus, making the expected overall score 0, thus, making game fair.

Values of  X = -2(if lost, or say chose any four colored marble except blue marble), or p(if won, or say chose blue colored marble)

• P(X = -2) = P(choosing non-blue marble) = 4/5
• P(X = p) = P(choosing blue marble) = 1/5

(its because choosing one marble from 5 marbles can be done in 5 ways, so n(S) = 5, and choosing 1 blue marble out of 5 different colored marble can be done in 1 way, and choosing 1 non-blue marble out of 5 different colored marble can be done in 4 ways)

Thus, we get:

$E(X) = -2 \times P(X = -2) + p \times P(X = p)\\E(X) = -2 \times 4/5 + p \times 1/5\\0 = -8/5 + p/5\\0 = -8 + p\\p = 8$

Thus, the points the blue marble should be worth for the game to be fair is given by: Option C: 8

Learn more about expectation of a random variable here:

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