Question

Nicole has a deck of 10 cards numbered 1 through 10. She is playing a game of chance. This game is this: Nicole chooses one card from the deck at random. She wins an amount of money equal to the value of the eard if an odd numbered card is drawn. She loses $3.40 if an even numbered card is drawn. (If necessary, consult a list of formulas.) (a) Find the expected value of playing the game. dofiars (b) What can Nicole expect in the long run, after playing the game many times? (She replaces the card in the deck each time.) Nicoie can expect to gain money. She can expect to win dollars per draw. Nicole can expect to lose money. She can expect to lose dollars per draw. Nicele can expect to break even (neither gain nor lose money).

Answer 1

Nicole can expect to gain money in the long run with this game of chance, averaging a win of 80 cents per draw due to a positive expected value from the mix of odd and even card outcomes.

Step-by-step explanation:

Nicole has a deck of 10 cards numbered 1 through 10. She is playing a game of chance where drawing an odd-numbered card yields money equal to the card's value, whereas drawing an even-numbered card results in a loss of $3.40.

First, let's calculate the expected value of playing the game. There are five odd-numbered cards (1, 3, 5, 7, 9) and five even-numbered cards (2, 4, 6, 8, 10).

The probability of drawing an odd card is 5/10 or 1/2, and the probability of drawing an even card is also 1/2. The expected win/loss for each event is calculated:

• For odd cards: (1 + 3 + 5 + 7 + 9)/5 = 5
• For even cards: -$3.40

Expected value = (1/2) * 5 + (1/2) * (-$3.40)

Expected value = $0.80

From this calculation, we conclude that in the long run, after playing the game many times, Nicole can expect to gain money. Specifically, she can expect to win 80 cents per draw on average. This is because the positive expected value indicates an excess of winning over losing.

Answer 2

(a) E = (1/10)($25) + (1/2)(-$3.40) = -$.80

(b) Nicole can expect to lose money.