SOLUTION: 1. In a game,using a fair die, a player bet #5 on a given number shows up, otherwise the #5 bet is won. What is the expected value of the game. 2. 5000 ticket is sold at #5 each f

Question 1197311: 1. In a game,using a fair die, a player bet #5 on a given number shows up, otherwise the #5 bet is won. What is the expected value of the game.
2. 5000 ticket is sold at #5 each for a raffle, ticket are to be drawn at random and price of #1000 each, 5 price of #200 each 20 price of #50 each. A man buy one of the raffle ticket. (a) what is the expected value for the man.
(b) calculate the probability that he win a price.
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**1. Expected Value of the Die Game**
* **Probability of Winning:** 1/6 (since there's one desired number on a 6-sided die)
* **Probability of Losing:** 5/6
* **Payoff for Winning:** $0 (bet is lost)
* **Payoff for Losing:** -$5 (bet amount)
* **Expected Value:** (Probability of Winning * Payoff for Winning) + (Probability of Losing * Payoff for Losing)
= (1/6) * $0 + (5/6) * (-$5)
= -$4.17
**Therefore, the expected value of the die game is -$4.17.** This means, on average, you're expected to lose $4.17 per game.
**2. Raffle Ticket Analysis**
* **Total Tickets Sold:** 5000
* **Prizes:**
* 1 prize of $1000
* 5 prizes of $200
* 20 prizes of $50
* **Total Prizes:** 1 + 5 + 20 = 26
* **(a) Probability of Winning a Prize:**
* Number of Winning Tickets / Total Tickets
* = 26 / 5000
* = 0.0052
* = 0.52%
* **(b) Expected Value for the Man:**
* **Total Prize Money:**
* ($1000 * 1) + ($200 * 5) + ($50 * 20) = $1000 + $1000 + $1000 = $3000
* **Expected Winnings:**
* (Total Prize Money / Total Tickets) - Cost of Ticket
* = ($3000 / 5000) - $5
* = $0.60 - $5
* = -$4.40
**Therefore, the expected value for the man is -$4.40.** This means, on average, he can expect to lose $4.40 per raffle ticket.

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