For the experiment described below, let \( x \) determine a random variable, and use your knowledge of probability to prepare a probability distribution. Three cards are drawn (with replacement) and the number of red cards is noted. A probability distribution for the given experiment is as follows. \( \mathrm{P}(\mathrm{x}) \)

is the probability of getting \( x \) red cards out of 3 draws. To construct the probability distribution, we need to consider the following: 1. The total number of possible outcomes when drawing 3 cards with replacement from a standard deck of 52 cards. 2. The number of ways to get exactly \( x \) red cards out of 3 draws. Let's start by calculating the total number of possible outcomes. Since each card is drawn with replacement, the probability of drawing a red card or a black card is the same for each draw. There are 26 red cards and 26 black cards in a standard deck of 52 cards. 1. Total number of possible outcomes: The total number of possible outcomes when drawing 3 cards with replacement is given by the formula for combinations with repetition (also known as the "stars and bars" method): \[ \binom{3 + 2}{2} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \] However, this calculation is incorrect. The correct total number of possible outcomes when drawing 3 cards with replacement is \( 52^3 \), since each draw has 52 possible outcomes (26 red and 26 black cards). 2. Number of ways to get exactly \( x \) red cards out of 3 draws: To get exactly \( x \) red cards out of 3 draws, we need to choose \( x \) red cards from the 26 available red cards and \( 3 - x \) black cards from the 26 available black cards. The number of ways to do this is given by the product of the combinations of choosing \( x \) red cards and \( 3 - x \) black cards: \[ \binom{26}{x} \cdot \binom{26}{3-x} \] Now, let's calculate the probability distribution for each possible value of \( x \) (0, 1, 2, 3) and sum the probabilities to ensure they add up to 1. 1. \( \mathrm{P}(x = 0) \): The probability of getting 0 red cards out of 3 draws. \[ \mathrm{P}(x = 0) = \frac{\binom{26}{0} \cdot \binom{26}{3}}{52^3} \] 2. \( \mathrm{P}(x = 1) \): The probability of getting 1 red card out of 3 draws. \[ \mathrm{P}(x = 1) = \frac{\binom{26}{1} \cdot \binom{26}{2}}{52^3} \] 3. \( \mathrm{P}(x = 2) \): The probability of getting 2 red cards out of 3 draws. \[ \mathrm{P}(x = 2) = \frac{\binom{26}{2} \cdot \binom{26}{1}}{52^3} \] 4. \( \mathrm{P}(x = 3) \): The probability of getting 3 red cards out of 3 draws. \[ \mathrm{P}(x = 3) = \frac{\binom{26}{3} \cdot \binom{26}{0}}{52^3} \] Now, let's calculate the probabilities for each value of \( x \): 1. \( \mathrm{P}(x = 0) \): \[ \mathrm{P}(x = 0) = \frac{\binom{26}{0} \cdot \binom{26}{3}}{52^3} = \frac{1 \cdot 2600}{140608} = \frac{2600}{140608} \] 2. \( \mathrm{P}(x = 1) \): \[ \mathrm{P}(x = 1) = \frac{\binom{26}{1} \cdot \binom{26}{2}}{52^3} = \frac{26 \cdot 325}{140608} = \frac{8300}{140608} \] 3. \( \mathrm{P}(x = 2) \): \[ \mathrm{P}(x = 2) = \frac{\binom{26}{2} \cdot \binom{26}{1}}{52^3} = \frac{325 \cdot 26}{140608} = \frac{8300}{140608} \] 4. \( \mathrm{P}(x = 3) \): \[ \mathrm{P}(x = 3) = \frac{\binom{26}{3} \cdot \binom{26}{0}}{52^3} = \frac{2600 \cdot 1}{140608