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.

A high school basketball team is selling $10 raffle tickets as part of a fund-raising program. The first prize is a trip to the Bahamas valued at $5460, and the second prize is a weekend ski package valued at $496. The remaining 18 prizes are $100 gas cards. The number of tickets sold is 3500.

Accepted Answer

Welcome back, everyone. In this problem, are charity is selling $15 raffle tickets to raise funds. The grand prize is a new laptop valued at $2200. The second prize is a bicycle worth $650 and there are 123 prizes of $120 gift cards. If 2000 tickets are sold, what is the expected value for a player who buys one ticket? A says it's about $13.72. In other words, the player will lose $13.72 per ticket bought. B says it's about $-12.86. C about $9.36 and D about $-8.48. Now, if we are going to figure out the expected value for a player who buys one ticket, we need to find the expected value for each event, for each outcome, OK? So, let's create a table. That shows the gain that is X for each outcome and then the probability, OK? P of X for each outcome. And now in this table, OK, we're going to do the gain. Let me put this here. OK. We're gonna do the gain for the laptop, the expected gain for the laptop. We're going to do it for the bicycle, OK. We're gonna do it for the uh the the the gift cards, OK, gift cards. And then of course we have to look at the game for the if if we won nothing or if the player wins nothing. Now how can we find each of these? Well, let's first start with the net gain for each outcome, OK? Now, when we think about the net gain, first, remember that we're spending $15 for the ticket. So for the laptop, the net gain is gonna be $2200 minus $15 which equals $2,185. OK? So that's $2,185. Now in the same spirit, that means for the bicycle, again, remember, it costs $650. So the net gain is $650 minus $15 or $635 OK. For the gift cards they cost $120. So it's gonna be uh. $120 minus $15 and that is $105 OK? And finally, if you win nothing, then you actually lose $15 so that's negative $15. Now what about the probabilities? Well, remember here we're told that 2000 tickets are sold and one laptop is going to be one. So remember the probability equals the number of successes divided by the total number of outcomes. So for a probability here, it's going to be 1 divided by 2000 for the laptop, the same for the bicycle because only one bicycle can be one. 3 divided by 2000 for the 12 divided by 1000 or 2000 for the gift cards because there are 12 prizes of gift cards, so that's 12 divided by 2000 or 3500. And then to win nothing, it would be the probability 1 minus the sum of the rest of the probabilities. OK? So that's really one. Let's make a note here for nothing to win nothing. The probability Would be equal to 1 minus the sum of the other probabilities. 12,000 plus 12,000 plus 3500. And that would be equal to 993,000, OK? So that's 993,000. Know that we have our expected values or expected gains and probabilities for each scenario, then we can calculate the expected value because no recall that expected value E is equal to the sum of values multiplied by their respective probabilities. So this would imply then that the expected value is going to be equal to 2,185 multiplied by 12,000 for the laptop plus 635 multiplied by 12,000 for the bicycle plus 105 multiplied by 3500 for the gift cards minus 15 multiplied by 993,100 if we want nothing. And in this case, when we calculate here and round it to the nearest cent we get approximately $12.86. In other words, on average, a player loses about $12.86 per ticket bought. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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Key Concepts

Expected Value

Probability

Game of Chance

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