Final Exam
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Question 35 of 40 (t point) I Question Attempt: 1 of 1
Lashonda has a deck of 10 cards numbered 1 through 10 . She is playing a game of chance.
This game is this: Lashonda chooses one card from the deck at random. She wins an amount of money equal to the value of the card if an odd numbered card is drawn. She loses $\$ 5$ if an even numbered card is drawn.
(a) Find the expected value of playing the game.
Dollars
(b) What can Lashonda expect in the long run, after playing the game many times?
(She replaces the card in the deck each time.)
Lashonda can expect to gain money.
She can expect to win $\square$ dollars per draw.
Lashonda can expect to lose money.
She can expect to lose $\square$ dollars per draw.
Lashonda can expect to break even (neither gain nor lose money).
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Step 1 ：The expected value is calculated by multiplying each outcome by its probability and then summing these products.

Step 2 ：There are 10 cards in total, 5 of which are odd and 5 of which are even. Therefore, the probability of drawing an odd card is \(\frac{5}{10} = 0.5\) and the probability of drawing an even card is also \(\frac{5}{10} = 0.5\).

Step 3 ：The odd cards are numbered 1, 3, 5, 7, 9. If Lashonda draws an odd card, she wins an amount of money equal to the value of the card. Therefore, the expected value from drawing an odd card is \((1*0.5 + 3*0.5 + 5*0.5 + 7*0.5 + 9*0.5) = 12.5\).

Step 4 ：If Lashonda draws an even card, she loses $5. Therefore, the expected value from drawing an even card is \(-5*0.5 = -2.5\).

Step 5 ：The total expected value of playing the game is the sum of these two expected values, which is \(12.5 - 2.5 = 10\).

Step 6 ：In the long run, Lashonda can expect to gain money. On average, she can expect to win $10 per draw.

Step 7 ：The final expected value of playing the game is \(\boxed{10}\).