Question

You and a friend are playing a game of chance. Every time you roll a 1 or 2 you are successful, and your friend will pay you $1. Every time you roll (using a fair die) a 3, 4, 5 or 6, you must pay your friend $2. If 71% of your rolls are successful and 29% of your rolls are unsuccessful, how much money do you expect to have earned/owed by the end of the game

Answer 1

Considering the mean of a discrete distribution, you are expected to earn $0.13 by the end of the game.

What is the mean of a discrete distribution?

The expected value of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

Since 71% of your rolls are successful and 29% of your rolls are unsuccessful, the distribution of your earnings is:

• P(X = 1) = 0.71.

• P(X = -2) = 0.29

Hence the expected value is:

E(X) = 1 x 0.71 - 2 x 0.29 = 0.71 - 0.58 = 0.13

More can be learned about the mean of a discrete distribution at brainly.com/question/24245882

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