91Ó°ÊÓ

First, we need to determine the probability of each winning outcome. A deck has 52 cards, 13 of which are hearts and 26 of which are black (spades or clubs).1. **Probability of drawing 3 hearts:** There are 13 hearts, and we draw 3 without replacement. The number of ways to choose 3 hearts from 13 is given by the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), which results in \[ C(13, 3) = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286 \] The number of ways to choose any 3 cards from a deck of 52 is \[ C(52, 3) = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100 \] So the probability of drawing 3 hearts is \[ P(3 \text{ hearts}) = \frac{286}{22100} \approx 0.01295 \]2. **Probability of drawing 3 black cards:** There are 26 black cards. Thus, the number of ways to choose 3 black cards is \[ C(26, 3) = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = 2600 \] So the probability of drawing 3 black cards is \[ P(3 \text{ black cards}) = \frac{2600}{22100} \approx 0.11765 \]3. **Probability of any other combination:** Since these are the only winning outcomes, the probability of a non-winning draw is: \[ P(\text{other}) = 1 - P(3 \text{ hearts}) - P(3 \text{ black cards}) \approx 1 - 0.01295 - 0.11765 = 0.8694 \]

Now, create a table for the probability model based on the outcomes and winnings:- Win \\(50 for 3 hearts: Probability \( P(X=50) = 0.01295 \)- Win \\)25 for 3 black cards: Probability \( P(X=25) = 0.11765 \)- Win \$0 for any other: Probability \( P(X=0) = 0.8694 \)

The expected winnings can be calculated using the formula for expected value:\[ E[X] = \sum (x_i \cdot P(x_i)) \]- For \\(50: \(0.01295 \times 50\)- For \\)25: \(0.11765 \times 25\)- For \\(0: \(0.8694 \times 0\)Calculate:\[ E[X] = (0.01295 \times 50) + (0.11765 \times 25) + (0.8694 \times 0) \approx 0.6475 + 2.94125 + 0 = 3.58875 \]The expected winnings are approximately \\)3.59.

First, calculate the variance using the formula:\[ \text{Var}(X) = E[X^2] - (E[X])^2 \]Calculate \(E[X^2]\):- For \\(50: \(0.01295 \times 50^2 = 32.375\)- For \\)25: \(0.11765 \times 25^2 = 73.53125\)- For \$0: \(0.8694 \times 0^2 = 0\)\[ E[X^2] = 32.375 + 73.53125 + 0 = 105.90625 \]Now, find the variance:\[ \text{Var}(X) = 105.90625 - (3.58875)^2 = 105.90625 - 12.8788 \approx 93.02745 \]Finally, the standard deviation is:\[ \text{SD}(X) = \sqrt{93.02745} \approx 9.646 \]

Subtract the cost of playing (\$5) from the expected winnings:- Expected net profit: \(E[X-5] = E[X] - 5 = 3.59 - 5 = -1.41\)The standard deviation remains the same as it is unaffected by constant shifts:- \(\text{SD}(X-5) = \text{SD}(X) = 9.646\)

Since the expected net profit is negative (\\(-1.41), you lose on average \\)1.41 per game. Considering the standard deviation, which suggests high variability, playing the game consistently results in a financial loss.