If x is the net 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 over the long run. A service organization is selling $2 raffle tickets as part of a fundraising program. The first prize is a boat valued at $2950, and the second prize is a camping tent valued at $400. In addition to the first and second prizes, there are twenty-four $23 gift certificates to be awarded. The number of tickets sold is 3000. Find the expected net gain to the player for one play of the game. Round your answer to the nearest cent.
A) $\$$ 1122.33
B) $\$$ 1.40
C) $\$$ 0.70
D)-$0.7
E)-$1122.33

Question

Quizplus · Accepted Answer

The expected net gain, E(x), for a player can be calculated using the formula for expected value, which is the sum of each outcome's value times its probability. Here, we calculate the expected value of winning each prize and subtract the cost of the ticket.- Probability of winning the boat: $ \frac{1}{3000} $- Expected value from the boat: $ \frac{2950}{3000} = \$0.9833 $- Probability of winning the camping tent: $ \frac{1}{3000} $- Expected value from the tent: $ \frac{400}{3000} = \$0.1333 $- Probability of winning a gift certificate: $ \frac{24}{3000} $- Expected value from a gift certificate: $ \frac{24 \times 23}{3000} = \$0.184 $Adding these values gives the total expected value from prizes: $ \$0.9833 + \$0.1333 + \$0.184 = \$1.3006 $Subtracting the cost of the ticket ($2) from the total expected value gives the expected net gain: $ \$1.3006 - \$2 = -\$0.6994 $, which rounds to -\$0.70.

If X Is the Net Gain to a Player in a Game