Another card game. In a new card game, you start with a well-shuffed full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.

(a) Create a probability model for the amount you win at this game, and nd the expected winnings. Also compute the standard deviation of this distribution.

My answer:
E(x)= 3.60
Standard deviation= 9.65

(b) If the game costs $5 to play, what would be the expected value and standard deviation of the net pro t (or loss)? (Hint: profit = winnings - cost; X - 5)

My answer:
3.60-5= -1.40

(c) If the game costs $5 to play, should you play this game? Explain.

My answer:
No.. because I would lose $1.40.

prob(3 hearts ) = C(13,3)/C(52/3)

= 286/22100
= 11/850

expectation = (11/850)($50) = $0.647

prob(3 black) = C(26,3)/C(52,3)
= 2600/22100
= 2/17

expectation = (2/7)($25) = $2.94

expected return on the game = $3.59

You were right on the $3.60 (difference is roundoff)

agree with the loss of $1.40

again, I don't know how sd is done in this kind of problem

Your answers for parts (a) and (b) are correct. However, your explanation for part (c) needs further clarification. Let's review the reasoning behind it.

In part (a), you correctly calculated the expected winnings to be $3.60 and the standard deviation to be $9.65. This means that, on average, you can expect to win $3.60 in this game. However, the standard deviation of $9.65 indicates that there is a lot of variability in the winnings.

In part (b), you correctly calculated the expected value of the net profit to be -$1.40. This means that, on average, you can expect to lose $1.40 per game when accounting for the $5 cost of playing. The standard deviation of the net profit would remain the same as the standard deviation of the winnings, which is $9.65.

In part (c), to determine whether you should play the game or not, you need to consider the expected net profit and the cost of playing the game. In this case, the expected net profit is negative, indicating that you are expected to lose money on average. Since the game costs $5 to play, you would be losing an additional $5 on top of the expected net profit of -$1.40. Therefore, it is not a favorable game to play, as you would end up losing money in the long run.

So the correct answer for part (c) is: No, you should not play this game because based on the expected net profit and the cost of playing, you are expected to lose money.

To create a probability model for the amount you win at this card game, we need to determine the probabilities of each possible outcome and the corresponding winnings.

Let's first consider the probability of drawing 3 hearts. There are 13 hearts in a deck of 52 cards, so the probability of drawing a heart on the first draw is 13/52. After the first draw, there are 12 hearts remaining in a deck of 51 cards, so the probability of drawing a heart on the second draw is 12/51. Similarly, the probability of drawing a heart on the third draw is 11/50. Since you must draw 3 hearts in a row to win, we multiply these probabilities together:

P(3 hearts) = (13/52) * (12/51) * (11/50) = 0.012

The winning amount for drawing 3 hearts is $50.

Next, let's consider the probability of drawing 3 black cards. There are 26 black cards (clubs and spades) in a deck of 52 cards, so the probability of drawing a black card on the first draw is 26/52. After the first draw, there are 25 black cards remaining in a deck of 51 cards, so the probability of drawing a black card on the second draw is 25/51. Similarly, the probability of drawing a black card on the third draw is 24/50. Multiplying these probabilities together:

P(3 black) = (26/52) * (25/51) * (24/50) = 0.1176

The winning amount for drawing 3 black cards is $25.

Lastly, the probability of any other draw (not 3 hearts or 3 black cards) is 1 - P(3 hearts) - P(3 black) = 1 - 0.012 - 0.1176 = 0.87

The winning amount for any other draw is $0.

Now, to find the expected winnings, we multiply each possible winning amount by its corresponding probability and sum them up:

E(x) = (0.012 * $50) + (0.1176 * $25) + (0.87 * $0) = $0.60 + $2.94 + $0 = $3.54

Therefore, the expected winnings for this game is $3.54.

To find the standard deviation of this distribution, we need to calculate the variance first. The variance (VAR(x)) is the sum of the squares of each possible deviation from the expected value (E(x)), multiplied by its probability:

VAR(x) = [(0 - $3.54)^2 * 0.87] + [($25 - $3.54)^2 * 0.1176] + [($50 - $3.54)^2 * 0.012]
= $0.0626 + $72.0474 + $579.8376
= $652.9476

The standard deviation (SD) is then the square root of the variance:

SD = sqrt($652.9476) ≈ $25.59

Now let's move on to part (b): if the game costs $5 to play, we need to calculate the net profit (or loss), which is the difference between the winnings and the cost:

Net profit = expected winnings - cost = $3.54 - $5 = -$1.46

Therefore, the expected value for the net profit is -$1.46.

The standard deviation for the net profit can be found by using the same formula as before. Since the cost is a fixed value, it does not affect the standard deviation calculation. Therefore, the standard deviation of the net profit remains the same as before: $25.59.

Finally, in part (c), whether you should play this game depends on your risk tolerance and personal preferences. In this case, the expected net profit is negative (-$1.46), meaning on average, you would lose $1.46 each time you play the game. If you are risk-averse and prefer not to lose money, it would be better to avoid playing this game. However, if you enjoy the thrill of playing and are willing to take risks, you might still choose to play despite the negative expected net profit.