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For each value of \(X\), we need to calculate the probability. We will do this by finding the different combinations of selecting red and green tents and dividing them by the total number of ways to select 4 tents from the 7. \(P(X=0)\): Karin selects no red tents, meaning she selects all green tents. There are \(\binom{4}{0}\binom{3}{4}\) ways to select 0 red tents and 4 green tents. The total number of possible selections is \(\binom{7}{4}\). Therefore, the probability is: \[P(X=0) = \frac{\binom{4}{0}\binom{3}{4}}{\binom{7}{4}} = 0\] \(P(X=1)\): Karin selects 1 red tent and 3 green tents. There are \(\binom{4}{1}\binom{3}{3}\) ways to select 1 red tent and 3 green tents. Therefore, the probability is: \[P(X=1) = \frac{\binom{4}{1}\binom{3}{3}}{\binom{7}{4}} = \frac{4}{35}\] \(P(X=2)\): Karin selects 2 red tents and 2 green tents. There are \(\binom{4}{2}\binom{3}{2}\) ways to select 2 red tents and 2 green tents. Therefore, the probability is: \[P(X=2) = \frac{\binom{4}{2}\binom{3}{2}}{\binom{7}{4}} = \frac{18}{35}\] \(P(X=3)\): Karin selects 3 red tents and 1 green tent. There are \(\binom{4}{3}\binom{3}{1}\) ways to select 3 red tents and 1 green tent. Therefore, the probability is: \[P(X=3) = \frac{\binom{4}{3}\binom{3}{1}}{\binom{7}{4}} = \frac{12}{35}\] \(P(X=4)\): Karin selects all 4 red tents. There are \(\binom{4}{4}\binom{3}{0}\) ways to select 4 red tents. Therefore, the probability is: \[P(X=4) = \frac{\binom{4}{4}\binom{3}{0}}{\binom{7}{4}} = \frac{1}{35}\] #Step 3: Write the probability distribution of X#

To find the expected number of red tents selected, we multiply the value of each \(X\) by its corresponding probability and sum the results: E[X] = 0 * P(X=0) + 1 * P(X=1) + 2 * P(X=2) + 3 * P(X=3) + 4 * P(X=4) E[X] = 0 * 0 + 1 * \(\frac{4}{35}\) + 2 * \(\frac{18}{35}\) + 3 * \(\frac{12}{35}\) + 4 * \(\frac{1}{35}\) E[X] = \(\frac{4 + 36 + 36 + 4}{35}\) E[X] = \(\frac{80}{35}\) So, on average, the expected number of red tents selected by Karin is \(\frac{80}{35}\).