Donna has a deck of 10 cards numbered 1 through 10 . She is playing a gaime of chancev
This game is this: Donna chooses one card from the deck at random. She wints an amount of money equal to the value of the card if an odd numbered card i
drawn. She loses if an eveh nimbered card is drawh.
(a) Find the expected value of playing the game.
II dollars
(b) What can Donna expect in the long run, after playing the game many times?
(She replaces the card in the deck each time.)
Donna can expect to gain money.
She can expect to win dollars per draw.
Donna can expect to lose money.
She can expect to lose dollars per draw.
Donna can expect to break even (neither gain nor lose money).

Ask by Gibson Hobbs. in the United States

Mar 30,2025

Question

Donna has a deck of 10 cards numbered 1 through 10 . She is playing a gaime of chancev
This game is this: Donna chooses one card from the deck at random. She wints an amount of money equal to the value of the card if an odd numbered card i
drawn. She loses if an eveh nimbered card is drawh.
(a) Find the expected value of playing the game.
II dollars
(b) What can Donna expect in the long run, after playing the game many times?
(She replaces the card in the deck each time.)
Donna can expect to gain money.
She can expect to win dollars per draw.
Donna can expect to lose money.
She can expect to lose dollars per draw.
Donna can expect to break even (neither gain nor lose money).

Ask by Gibson Hobbs. in the United States

Mar 30,2025

UpStudy AI · Accepted Answer

Donna's game involves drawing one card at random from a deck of 10 cards numbered 1 through 10. The payoff is defined as follows:
- If an odd-numbered card is drawn, she wins an amount equal to the value on the card.
- If an even-numbered card is drawn, she loses $$\$3.40$$.

### (a) Finding the Expected Value

1. **Determine the outcomes and their probabilities:**

There are 10 cards in total, so the probability for each card is $$\frac{1}{10}$$.

The odd-numbered cards are: $$1, 3, 5, 7, 9$$ (5 cards).

The even-numbered cards are: $$2, 4, 6, 8, 10$$ (5 cards).

2. **Calculate the contribution from the odd-numbered cards:**

For odd-numbered cards, the winnings equal the card's value. The expected contribution is:
   $$
   E_{\text{odd}} = \frac{1}{10}(1 + 3 + 5 + 7 + 9)
   $$
   Compute the sum:
   $$
   1 + 3 + 5 + 7 + 9 = 25
   $$
   So:
   $$
   E_{\text{odd}} = \frac{25}{10} = 2.5
   $$

3. **Calculate the contribution from the even-numbered cards:**

For even-numbered cards, Donna loses $$\$3.40$$ regardless of the number. The expected contribution is:
   $$
   E_{\text{even}} = \frac{1}{10}(-3.40) \times 5
   $$
   Simplify:
   $$
   E_{\text{even}} = \frac{-3.40 \times 5}{10} = \frac{-17.00}{10} = -1.70
   $$

4. **Combine the contributions to find the total expected value:**

$$
   E(X) = E_{\text{odd}} + E_{\text{even}} = 2.5 - 1.70 = 0.80
   $$

Thus, the expected value of playing the game is $$\$0.80$$.

### (b) Long-Run Expectation

Since Donna replaces the card each time (ensuring independence of draws), after playing the game many times, the average gain per draw will be equal to the expected value.

- Donna can expect to **win** money.
- She can expect to win $$\$0.80$$ per draw.
Donna can expect to win \$0.80 per draw.

Upstudy AI Solution

Answer

Solution

The Deep Dive

(a) Expected Value Calculation

(b) Long-Term Expectation

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