Finding an Expected Value In Exercises 37 and 38, find the exp...

Question

Finding an Expected Value In Exercises 37 and 38, find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose.

In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, . . ., 34, 35, and 36, marked on equally spaced slots. If a player bets $1 on a number and wins, then the player keeps the dollar and receives an additional $35. Otherwise, the dollar is lost.

Accepted Answer

Welcome back, everyone. In a lottery game, there are 100 tickets sold, each costing $$2. 1 ticket is drawn at random as the winner. The winner receives a price of 150 $$which includes the cost of the ticket. Find the expected value for a player who buys one ticket. A negative $$0.85 B negative 0.75 C $$negative $$0.50 and D 0.25. So $$for this problem, we want to identify the expected value. Let's recall that in order to identify the expected value E of X, we simply have to sum the probabilities multiplied by their corresponding. XI values, right? So essentially we're going to multiply by the random variable values. What we have to do is simply construct the table for simplicity. So, in our first column, we can say that we are representing our net profit. Which is X, right? Our random variable, and in the second column, we're going to define the corresponding probabilities. In other words, that would be P of X. So now what are the possibilities for the net profit? Well, we have two possible scenarios, right? We either win or lose and the ticket costs $$2. So in order to identify the possible net profit values, we first have to understand that we can gain. Negative 2 $$that would be our net profit if we lose, right? Because we're investing $$2 and basically we're losing them. So we can have negative 2 or we $$can win $$150. But we also have to subtract our initial investment of 2 $$which gives us a net profit of $$148. So that would be our second possibility, 148 of $$net profit. So now we have our net profit values and we want to define the corresponding probabilities. Now there are 100 tickets sold, right? And one ticket is going to be the winning ticket. So, the probability of a when, let's call it the probability of W, is the number of favorable outcomes, which is 1 divided by the total number of outcomes, which is the number of tickets, and that's 100. And the probability of losing, meaning not winning is basically the complement of that, right? 1 minus the probability of winning, which is 99 divided by 100. So our probabilities are 99 divided by 100. For losing and 1 divided by 100 for winning. So now we can use this table and apply the formula to identify the expected value. We have to multiply. Our first probability 99 divided by 100. By the net profit of $$2 and then we're going to add our probability of winning one divided by 100. Which is multiplied by the net profit of 148. $$When we perform the calculation, we end up with negative $0.50. Which corresponds to the answer choice C. Thank you for watching.

Key Concepts

Expected Value

Probability

Game of Chance

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