#32 ### Calculating the Expected Value of a Card GameIn this exercise, we will determine the expected value for a game involving a standard 52-card deck. This will help us understand how much a player can expect to win or lose on average per game.#### Game Rules:1. A player must pay $14 to participate in the game.2. The goal is to draw a card from a standard 52-card deck.3. If a club is drawn (one of the 13 clubs in the deck), the player wins $21.4. If any other suit is drawn (one of the 39 non-club cards), the player loses their $14.#### Problem Statement:Calculate the price that would make the game fair for the player.#### Calculations:1. **Determine the Probability of Each Outcome:** - Probability of drawing a club (P(club)): \[ \frac{13}{52} = \frac{1}{4} = 0.25 \] - Probability of drawing a non-club (P(non-club)): \[ \frac{39}{52} = \frac{3}{4} = 0.75 \]2. **Calculate the Winnings and Losses:** - Winning amount when a club is drawn: $21 - Loss amount when a non-club is drawn: $143. **Expected Value Calculation:** The expected value (EV) is given by the formula: \[ \text{EV} = (\text{P(club)} \times \text{Winnings}) + (\text{P(non-club)} \times \text{Loss}) \] Substituting the probabilities and amounts: \[ \text{EV} = (0.25 \times 21) + (0.75 \times (-14)) \] \[ \text{EV} = (5.25) + (-10.5) \] \[ \text{EV} = -5.25 \]4. **Fair Game Calculation:** For the game to be fair, the expected value should be zero. Let "X" be the price that makes the game fair: \[ 0 = (0.25 \times 21) + (0.75 \

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to play the game, which must be subtracted from the winnings. If a club is drawn, the player wins 21 dollars; otherwis they lose their 14 dollars. Calculate the price that would make the game fair. The expected value of this game is dollars. (Round to the nearest cent as needed.)